NCERT Solutions for Class 8 Maths Chapter 14 Factorisation | एनसीईआरटी कक्षा 8 गणित अध्याय 14 गुणनखंड के समाधान class 8th maths
NCERT Solutions for Class 8 Maths Chapter 14 Factorisation class 8th maths गुणनखंड। एनसीइआरटी कक्षा 8 गणित प्रश्नावली 14 गुणनखंड के सभी प्रश्न उत्तर सवालों के जवाब सम्मिलित है। Here We learn what is in ncert maths class 8 chapter 14 गुणनखंड and how to solve questions with easiest method. class 8 maths chapter 14 In this chapter we solve the question of NCERT class 8 maths chapter 14.
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NCERT Solutions for Class 8 Maths Chapter 14 Factorisation class 8th maths
Ex 14
प्रश्नावली – 14
गुणनखंड
गणित
NCERT SOLUTIONS FOR CLASS 8 MATHS CHAPTER 14
ncert solutions for class 8 maths class 8th maths
Ex. 14.1
प्रश्नावली 14.1
1. दिए हुए पदों में सार्व गुणनखंड ज्ञात कीजिए :
(i) 12?, 36 (ii) 2?, 22?? (iii) 14??, 28?²?²
(iv) 2?, 3?², 4 (v) 6???, 24??², 12?²?
(vi) 16?³, -4?², 32? (vii) 10??, 20??, 30??
(viii) 3?²?³, 10?³?², ?²?²?
हल : (i) 12?, 36
12? = 2 × 2 × 3 × ?
36 = 2 × 2 × 3 × 3
सार्व गुणनखंड = 2 × 2 × 3 = 12
(ii) 2?, 22??
2y = 2 × y
22xy = 2 × 11 × x × y
सार्व गुणनखंड = 2
(iii) 14pq, 28p²q²
14pq = 2 × 7 × p × q
28p²q² = 2 × 2 × 7 × p × p × q × q
सार्व गुणनखंड = 2 × 7 × p × q = 14pq
(iv) 2x, 3x², 4
2x = 2 × x
3x² = 3 × x × x
4 = 2 × 2
चूँकि तीनों पदों में कोई भी उभयनिष्ठ नहीं है अतः सार्व गुणनखंड = 1
(v) 6abc, 24ab², 12a²b
6abc = 2 × 3 × a × b × c
24ab² = 2 × 2 × 2 × 3 × a × b × b
12a²b = 2 × 2 × 3 × a × a × b
सार्व गुणनखंड = 2 × 3 × b = 6b
(vi) 16x³, -4x², 32x
16x³ = 2 × 2 × 2 × 2 × x × x × x
− 4x² = − 2 × 2 × x × x
32x = 2 × 2 × 2 × 2 × 2 × x
सार्व गुणनखंड = 2 × 2 × x = 4x
(vii) 10pq, 20qr, 30rp
10pq = 2 × 5 × p × q
20qr = 2 × 2 × 5 × q × r
30rp = 2 × 3 × 5 × r × p
सार्व गुणनखंड = 2 × 5 = 10
(viii) 3x²y³, 10x³y², x²y²z
3x²y³ = 3 × x × x × y × y × y
10x³y² = 2 × 5 × x × x × x y × y
x²y²z = x × x × y × y × z
सार्व गुणनखंड = x × x × y × y = x²y²
2. निम्नलिखित व्यंजकों के गुणनखंड कीजिए :
NCERT SOLUTIONS FOR CLASS 8 MATHS CHAPTER 8
(i) 7x – 42 (ii) 6p – 12q (iii) 7a² + 14a
(iv) – 16z + 20z (v) 20l²m + 30alm
(vi) 5x²y – 15xy² (vii) 10a² – 15b² + 20c²
(viii) – 4a² + 4ab – 4ca (ix) x²yz + xy²z + xyz²
(x) ax²y + bxy² + cxyz
हल :
(i) 7x – 42
दोनों पदों के गुणनखंड करने पर –
7x = 7 × x
42 = 2 × 3 7
दोनों पदों में 7 सार्व गुणनखंड है
अतः 7x – 42
= (7 × x) – (7 × 2 × 3)
= 7 × [x – (2 × 3)]
= 7 (x – 6)
(ii) 6p – 12q
दोनों पदों के गुणनखंड करने पर –
6p = 2 × 3 × p
12q = 2 × 2 × 3 × q
दोनों पदों में 2 और 3 सार्व गुणनखंड है।
अतः 6p – 12q
= (2 × 3 × p) – (2 × 3 × 2 × q)
= 2 × 3 × [p – (2 × q)]
= 6 (p – 2q)
(iii) 7a² + 14a
दोनों पदों के गुणनखंड करने पर –
7a² = 7 × a × a
14a = 2 × 7 × a
दोनों पदों में 7 और a सार्व गुणनखंड है।
अतः 7a² + 14a
= (7 × a × a) + (7 × a × 2)
= 7 × a × (a + 2)
= 14 (a + 2)
(iv) – 16z + 20z²
दोनों पदों के गुणनखंड करने पर –
– 16z = – 2 × 2 × 2 × 2 × z
20z² = 2 × 2 × 5 × z × z
दोनों पदों में 2, 2 और z सार्व गुणनखंड है।
अतः – 16z + 20z²
= (- 2 × 2 × 2 × 2 × z) + (2 × 2 × z × 5 × z)
= 2 × 2 × z × [(- 2 × 2) + (5 × z)]
= 4z (-4 + 5z)
(v) 20l²m + 30alm
दोनों पदों के गुणनखंड करने पर –
20l²m = 2 × 2 × 5 × l × l × m
30alm = 2 × 3 × 5 × a × l × m
दोनों पदों में 2, 5, l और m सार्व गुणनखंड है।
अतः 20l²m + 30alm
= (2 × 5 × l × m × 2 × l) + ((2 × 5 × l × m × a)
= 2 × 5 × l × m × [(2 × l) + a]
= 10lm (2l + a)
(vi) 5x²y – 15xy²
दोनों पदों के गुणनखंड करने पर –
5x²y = 5 × x × x × y
15xy² = 3 × 5 × x × y × y
दोनों पदों में 5, x और y सार्व गुणनखंड है।
अतः 5x²y – 15xy²
= (5 × x × y × x) – (5 × x × y × 3 × y)
= 5 × x × y × [x – (3 × y)]
= 5xy (x – 3y)
(vii) 10a² – 15b² + 20c²
तीनों पदों के गुणनखंड करने पर –
10a² = 2 × 5 × a × a
15b² = 3 × 5 × b × b
20c² = 2 × 2 × 5 × c × c
तीनों पदों में 5 सार्व गुणनखंड है।
अतः 10a² – 15b² + 20c²
= (5 × 2 × a × a) – (5 × 3 × b × b) + (5 × 2 × 2 × c × c)
= 5 × [(2 × a × a) – (3 × b × b) + (2 × 2 × c × c)]
= 5 (2a² – 3b² + 4c²)
(viii) – 4a² + 4ab – 4ca
तीनों पदों के गुणनखंड करने पर –
– 4a² = – 2 × 2 × a × a
4ab = 2 × 2 × a × b
– 4ca = 2 × 2 × c × a
तीनों पदों में 2, 2 और a सार्व गुणनखंड है।
अतः – 4a2 + 4ab – 4ca
= – (2 × 2 × a × a) + (2 × 2 × a × b) – (2 × 2 × a × c)
= 2 × 2 × a × (- a + b – c)
= 4a (-a + b + c)
(ix) x²yz + xy²z + xyz²
तीनों पदों के गुणनखंड करने पर –
x²yz = x × x × y × z
xy²z = x × y × y × z
xyz² = x × y × z × z
तीनों पदों में x, y और z सार्व गुणनखंड है।
अतः x²yz + xy²z + xyz²
= (x × y × z × x) + (x × y × z × y) + (x × y × z × z)
= x × y × z × (x + y + z)
= xyz (x + y + z)
(x) ax²y + bxy² + cxyz
तीनों पदों के गुणनखंड करने पर –
ax²yz = a × x × x × y
bxy² = b × x × y × y
cxyz = c × x × y × z
तीनों पदों में x और y सार्व गुणनखंड है।
अतः ax²y + bxy² + cxyz
= (x × y × a × x) + (x × y × b × y) + (x × y × c × z)
= x × y × [(a × x) + (b × y) + (c × z)]
= xy (ax + by + cz)
3. गुणनखंड कीजिए :
(i) x² + xy + 8x + 8y (ii) 15xy -6x + 5y – 2
(iii) ax + bx – ay – by (iv) 15pq + 15 + 9q + 25p
(v) z – 7 + 7xy – xyz
हल :
(i) x² + xy + 8x + 8y
सभी पदों में कोई भी सार्व गुणनखंड नहीं है, अतः पुनः समूहन करने पर –
= (x² + xy) + (8x + 8y)
= x (x + y) + 8 (x + y)
= (x + y) (x + 8)
(ii) 15xy -6x + 5y – 2
सभी पदों में कोई भी सार्व गुणनखंड नहीं है, अतः पुनः समूहन करने पर –
= 15xy + 5y -6x – 2
= 5y (3x + 1) – 2 (3x – 2)
= (5y – 2) (5y – 2)
(iii) ax + bx – ay – by
सभी पदों में कोई भी सार्व गुणनखंड नहीं है, अतः पुनः समूहन करने पर –
= (ax + bx) – (ay – by)
= x (a + b) – y (a + b)
= (a + b) (x – y)
(iv) 15pq + 15 + 9q + 25p
सभी पदों में कोई भी सार्व गुणनखंड नहीं है, अतः पुनः समूहन करने पर –
= 15pq + 15 + 9q + 25p
= (15pq + 25p) + (9q + 15)
= 5p (3q + 5) + 3 (3q + 5)
= (5p + 3) +(5p + 3)
(v) z – 7 + 7xy – xyz
सभी पदों में कोई भी सार्व गुणनखंड नहीं है, अतः पुनः समूहन करने पर –
= z – 7 + 7xy – xyz
= z – xyz – 7 + 7xy
= z (1 – xy) – 7 (1 – xy)
= (z – 7) (1 – xy)
NCERT SOLUTIONS FOR CLASS 8 MATHS CHAPTER 8
ncert solutions for class 8 maths class 8th maths
Ex. 14.2
प्रश्नावली 14.2
1. निम्नलिखित व्यंजकों के गुणनखंड कीजिए :
(i) a² + 8a +16 (ii) p² – 10p + 25 (iii) 25m² + 30m + 9
(iv) 49y² + 84yz + 36z² (v) 4x² – 8x + 4
(vi) 121b² – 88bc + 16c²
(vii) (l + m)² – 4lm (संकेत : पहले (l + m)² को प्रसारित कीजिए।)
(viii) a⁴ + 2a²b² + b⁴
हल : (i) a² + 8a +16
= a² + 4a + 4a + 16
= a (a + 4) + 4 (a + 4)
= (a + 4) (a + 4)
∵ (a + b) (a + b) = (a + b)²
∴ (a +4)²
(ii) p² – 10p + 25
= p² – 5p – 5p + 25
= p (p – 5) – 5 (p – 5)
= (p – 5) (p – 5)
∵ (a – b) (a – b) = (a – b)²
∴ (p – 5)²
(iii) 25m² + 30m + 9
= 25m² + 15m + 15m + 9
= 5m (5m + 3) + 3 (5m + 3)
= (5m + 3) (5m + 3)
∵ (a + b) (a + b) = (a + b)²
∴ (5m + 2)²
(iv) 49y² + 84yz + 36z²
= 49y² + 42yz + 42yz + 36z²
= 7y (7y + 6z) + 6z (7y + 6z)
= (7y +6z) + (7y + 6z)
∵ (a + b) (a + b) = (a + b)²
∴ (7y + 6z)²
(vi) 121b² – 88bc + 16c²
= 121b² – 44bc – 44bc + 16c²
= 11b (11b – 4c) – 4c (11b – 4c)
= (11b – 4c) (11b – 4c)
∵ (a + b) (a + b) = (a + b)²
∴ (11b – 4c)²
(vii) (l + m)² – 4lm
∵ (a + b)² = a² + 2ab + b²
∴ l² + 2lm + m² – 4lm
= l² + 2lm – 4lm + m²
= l² – 2lm + m²
∵ a² – 2ab + b² = (a – b)²
∴ (l – m)²
(viii) a⁴ + 2a²b² + b⁴
= (a²)² + 2a²b² + (b²)²
∵ a² + 2ab + b² = (a + b)²
∴ (a² + b²)²
2. गुणनखंड कीजिए :
(i) 4p² – 9q² (ii) 63a² – 112b² (iii) 49x² – 36
(iv) 16x⁵ – 144x³ (v) (l + m)² – (l – m)²
(vi) 9x²y² – 16 (vii) (x² – 2xy + y²) – z²
(viii) 25a² – 4b² + 28bc – 49c²
हल : (i) 4p² – 9q²
= (2p)² – (3q)²
∵ a² – b² = (a + b) (a – b)
= (2p + 3q) (2p – 3q)
(ii) 63a² – 112b²
= 7 (9a² – 16b²)
= 7 [(3a)² – (4b)²]
∵ a² – b² = (a + b) (a – b)
= 7 (3a + 4b) (3a – 4b)
(iii) 49x² – 36
= (7x)² – (6)²
∵ a² – b² = (a + b) (a – b)
= (7x + 6) (7x -6)
(iv) 16x⁵ – 144x³
= 16x³ (x² – 9)
= 16x³ (x² – 3²)
∵ a² – b² = (a + b) (a – b)
= 16x³ (x + 3) (x – 3)
(v) (l + m)² – (l – m)²
∵ a² – b² = (a + b) (a – b)
= (l + m + l – m) [(l + m) – ( l – m)]
= 2l (l + m – l + m)
= 2l × 2m
= 4lm
(vi) 9x²y² – 16
= (3xy)² – (4)²
∵ a² – b² = (a + b) (a – b)
= (3xy + 4) (3xy – 4)
(vii) (x² – 2xy + y²) – z²
∵ a² – 2ab + b² = (a – b)²
= (x – y)² – z²
∵ a² – b² = (a + b) (a – b)
= (x – y + z) (x – y – z)
(viii) 25a² – 4b² + 28bc – 49c²
पुनः समूहन करने पर –
= 25a² – (4b² – 28bc + 49c²)
= 25a² – [(2b)² – 2 × 2b × 7c + (7c)²]
∵ a² – 2ab + b² = (a – b)²
= 25a² – (2b – 7c)²
∵ a² – b² = (a + b) (a – b)
= (5a)² – (2b – 7c)²
= (5a + 2b – 7c) (5a – 2b – 7c)
3. निम्नलिखित व्यंजकों के गुणनखंड कीजिए :
(i) ax² + bx (ii) 7p² + 21q² (iii) 2x³ + 2xy² + 2xz²
(iv) am² + bm² + bn² + an² (v) (lm + l) + m + 1
(vi) y (y + z) + 9 (y + z) (vii) 5y² – 20y – 8z + 2yz
(viii) 10ab + 4a + 5b +2 (ix) 6xy – 4y + 6 – 9x
हल : (i) ax² + bx
= x = (ax + b)
(ii) 7p² + 21q²
= 7 (p² + 3q²)
(iii) 2x³ + 2xy² + 2xz²
= 2x (x² + y² + z²)
(iv) am² + bm² + bn² + an²
पुनः समूहन करने पर –
= (am² + bm²) + (an² + bn²)
= m² (a + b) + n² (a + b)
= (a + b) (m² + n²)
(v) (lm + l) + m + 1
पुनः समूहन करने पर –
= (lm + m) + (l + 1)
= m (l + 1) + 1 (l + 1)
= (l + 1) (m + 1)
(vi) y (y + z) + 9 (y + z)
= (y + z) (y + 9)
(vii) 5y² – 20y – 8z + 2yz
पुनः समूहन करने पर –
= 5y² – 20y + 2yz – 8z
= 5y (y – 4) + 2z (y – 4)
= (y – 4) (5y + 2z)
(viii) 10ab + 4a + 5b + 2
पुनः समूहन करने पर –
= 10ab + 5b + 4a + 2
= 5b (2a + 1) + 2 (2a + 1)
= (2a + 1) (5b + 2)
(ix) 6xy – 4y + 6 – 9x
पुनः समूहन करने पर –
= 6xy – 4y -9x + 6
= 2y (3x – 2) – 3 (3x + 2)
= (3x – 2) (2y -3)
4. गुणनखंड कीजिए :
NCERT SOLUTIONS FOR CLASS 8 MATHS CHAPTER 8
(i) a⁴ – b⁴ (ii) p⁴ – 81 (iii) x⁴ – (y + z)⁴
(iv) x⁴ – (x – z)⁴ (v) a⁴ – 2a²b² + b⁴
हल : (i) a⁴ – b⁴
= (a²)² – (b²)²
chunki a² – b² = (a + b) (a – b)
= (a² + b²) (a² – b²)
= (a² + b²) (a + b) (a – b)
(ii) p⁴ – 81
= (p²)² – (3²)²
chunki a² – b² = (a + b) (a – b)
= (p² + 3²) (p² – 3²)
= (p² + 3²) (p + 3) (p – 3)
(iii) x⁴ – (y + z)⁴
= (x²)² – [(y + z)²]²
= [x² + (y + z)²] + [x² – (y + z)²]
= [x² + (y + z)²] + (x + y + z) [x – (y + z)]
= [x² + (y + z)²] + (x + y + z) (x – y – z)
(v) a⁴ – 2a²b² + b⁴
= (a²)² – 2 × a² × b² + (b²)²
chunki a² – 2ab + b² = (a – b)²
= (a² – b²)²
5. निम्नलिखित व्यंजकों के गुणनखंड कीजिए :
(i) p² + 6p + 8 (ii) q² – 10q +21 (iii) p² + 6p – 16
हल : (i) p² + 6p + 8
= p² + 4p + 2p + 8
= p (p + 4) + 2 (p + 4)
= (p + 4) (p + 2)
(ii) q² – 10q +21
= q² – 7q – 3q +21
= q (q – 7) – 3 (q – 7)
= (q – 7) (q – 3)
(iii) p² + 6p – 16
= p² + 8p – 2p – 16
= p (p + 8) – 2 (p + 8)
= (p + 8) (p – 2)
NCERT SOLUTIONS FOR CLASS 8 MATHS CHAPTER 8
ncert solutions for class 8 maths class 8th maths
Ex. 14.3
प्रश्नावली 14.3
1. निम्नलिखित विभाजन कीजिए :
(i) 28x⁴ ÷ 56x (ii) – 36y³ ÷ 9y² (iii) 66pq²r³ ÷ 11 qr²
(iv) 34x³y³z³ ÷ 51xy²z³ (v) 12a⁸b⁸ ÷ (-6a⁶b⁴)
हल :
(i) 28x⁴ ÷ 56x
\(\displaystyle \begin{array}{l}=\frac{{\left( {2\times 2\times 7\times x\times x\times x\times x} \right)}}{{\left( {2\times 2\times 2\times 7\times x} \right)}}\\=\frac{{{{x}^{3}}}}{2}\end{array}\)
\(\displaystyle \begin{array}{l}=\frac{{\left( {2\times 2\times 7\times x\times x\times x\times x} \right)}}{{\left( {2\times 2\times 2\times 7\times x} \right)}}\\=\frac{{{{x}^{3}}}}{2}\end{array}\)
(ii) – 36y³ ÷ 9y²
\(\displaystyle =\frac{{-\left( {2\times 2\times 3\times 3\times y\times y\times y} \right)}}{{\left( {3\times 3\times y\times y} \right)}}\)
= − 2 × 2 × y
= − 4y
(iii) 66pq²r³ ÷ 11 qr²
\(\displaystyle =\frac{{2\times 3\times 11\times p\times q\times q\times r\times r\times r}}{{11\times q\times r\times r}}\)
= 2 × 3 × p × q × r
= 6pqr
(iv) 34x³y³z³ ÷ 51xy²z³
\(\displaystyle =\frac{{2\times 17\times x\times x\times x\times y\times y\times y\times z\times z\times z}}{{3\times 17\times x\times y\times y\times z\times z\times z}}\)\(\displaystyle =\frac{{2\times x\times x\times y}}{3}\)
\(\displaystyle =\frac{2}{3}{{x}^{2}}y\)
\(\displaystyle =\frac{{2\times 17\times x\times x\times x\times y\times y\times y\times z\times z\times z}}{{3\times 17\times x\times y\times y\times z\times z\times z}}\)\(\displaystyle =\frac{{2\times x\times x\times y}}{3}\)
\(\displaystyle =\frac{2}{3}{{x}^{2}}y\)
(v) 12a⁸b⁸ ÷ (-6a⁶b⁴)
\(\displaystyle =\frac{{2\times 2\times 3\times a\times a\times a\times a\times a\times a\times a\times a\times b\times b\times b\times b\times b\times b\times b\times b}}{{-2\times 3\times a\times a\times a\times a\times a\times a\times b\times b\times b\times b\times b\times b}}\)
\(\displaystyle =\frac{{2\times 2\times 3\times a\times a\times a\times a\times a\times a\times a\times a\times b\times b\times b\times b\times b\times b\times b\times b}}{{-2\times 3\times a\times a\times a\times a\times a\times a\times b\times b\times b\times b\times b\times b}}\)
= – 2 × a × a × a × a × b × b
= – 2a²b⁴
2. दिए हुए बहुपद को दिए हुए एकपदी से भाग दीजिए :
(i) (5x² – 6x) ÷ 3x (ii) (3y⁸ – 4y⁶ + 5y⁴) ÷ y⁴
(iii) 8 (x³y²z² + x²y³z² + x²y²z³) ÷ 4x²y²z²
(iv) (x³ + 2x² + 3x) ÷ 2x (v) (p³q⁶ – p⁶q³) ÷ p³q³
हल :
(i) (5x² – 6x) ÷ 3x
\(\displaystyle =\frac{{x\left( {5x-6} \right)}}{{3x}}\)
\(\displaystyle =\frac{{\left( {5x-6} \right)}}{3}\)
\(\displaystyle =\frac{1}{3}\left( {5x-6} \right)\)
\(\displaystyle =\frac{{x\left( {5x-6} \right)}}{{3x}}\)
\(\displaystyle =\frac{{\left( {5x-6} \right)}}{3}\)
\(\displaystyle =\frac{1}{3}\left( {5x-6} \right)\)
(ii) (3y⁸ – 4y⁶ + 5y⁴) ÷ y⁴
\(\displaystyle =\frac{{\left( {3{{y}^{8}}-4{{y}^{6}}+5{{y}^{4}}} \right)}}{{{{y}^{4}}}}\)\(\displaystyle =\frac{{{{y}^{4}}\left( {3{{y}^{4}}-4{{y}^{2}}+5} \right)}}{{{{y}^{4}}}}\)
\(\displaystyle =\frac{{\left( {3{{y}^{8}}-4{{y}^{6}}+5{{y}^{4}}} \right)}}{{{{y}^{4}}}}\)\(\displaystyle =\frac{{{{y}^{4}}\left( {3{{y}^{4}}-4{{y}^{2}}+5} \right)}}{{{{y}^{4}}}}\)
= (3y⁴ − 4y² +5)
(iii) 8 (x³y²z² + x²y³z² + x²y²z³) ÷ 4x²y²z²
\(\displaystyle =\frac{{8\left( {{{x}^{3}}{{y}^{2}}{{z}^{2}}+{{x}^{2}}{{y}^{3}}{{z}^{2}}+{{x}^{2}}{{y}^{2}}{{z}^{3}}} \right)}}{{4{{x}^{2}}{{y}^{2}}{{z}^{2}}}}\)
\(\displaystyle =\frac{{8{{x}^{2}}{{y}^{2}}{{z}^{2}}\left( {x+y+z} \right)}}{{4{{x}^{2}}{{y}^{2}}{{z}^{2}}}}\)
= 2 (x + y + z)
\(\displaystyle =\frac{{8\left( {{{x}^{3}}{{y}^{2}}{{z}^{2}}+{{x}^{2}}{{y}^{3}}{{z}^{2}}+{{x}^{2}}{{y}^{2}}{{z}^{3}}} \right)}}{{4{{x}^{2}}{{y}^{2}}{{z}^{2}}}}\)
\(\displaystyle =\frac{{8{{x}^{2}}{{y}^{2}}{{z}^{2}}\left( {x+y+z} \right)}}{{4{{x}^{2}}{{y}^{2}}{{z}^{2}}}}\)
= 2 (x + y + z)
(iv) (x² + 2x² + 3x) ÷ 2x
\(\displaystyle \begin{array}{l}=\frac{{{{x}^{2}}+2{{x}^{2}}+3x}}{{2x}}\\=\frac{{x\left( {x+2x+3} \right)}}{{2x}}\\=\frac{1}{2}\left( {x+2x+3} \right)\end{array}\)
\(\displaystyle \begin{array}{l}=\frac{{{{x}^{2}}+2{{x}^{2}}+3x}}{{2x}}\\=\frac{{x\left( {x+2x+3} \right)}}{{2x}}\\=\frac{1}{2}\left( {x+2x+3} \right)\end{array}\)
(v) (p³q⁶ – p⁶q³) ÷ p³q³
\(\displaystyle \begin{array}{l}=\frac{{{{p}^{3}}{{q}^{6}}-{{p}^{6}}{{q}^{3}}}}{{{{p}^{3}}{{q}^{3}}}}\\=\frac{{{{p}^{3}}{{q}^{3}}\left( {{{q}^{3}}-{{p}^{3}}} \right)}}{{{{p}^{3}}{{q}^{3}}}}\end{array}\)
\(\displaystyle \begin{array}{l}=\frac{{{{p}^{3}}{{q}^{6}}-{{p}^{6}}{{q}^{3}}}}{{{{p}^{3}}{{q}^{3}}}}\\=\frac{{{{p}^{3}}{{q}^{3}}\left( {{{q}^{3}}-{{p}^{3}}} \right)}}{{{{p}^{3}}{{q}^{3}}}}\end{array}\)
= q³ – p³
3. निम्नलिखित विभाजन कीजिए :
(i) (10x – 25) ÷ 5 (ii) (10x – 25) ÷ (2x – 5)
(iii) 10y (6y + 21) ÷ 5 (2y +7)
(iv) 9x²y² (3z -24) ÷ 27xy (z-8)
(v) 96abc (3a – 12) (5b – 30) ÷ 144 (a – 4) (b – 6)
हल :
(i) (10x – 25) ÷ 5
\(\displaystyle \begin{array}{l}=\frac{{\left( {10x-25} \right)}}{5}\\=\frac{{5\left( {2x-5} \right)}}{5}\end{array}\)
= (2x − 5)
\(\displaystyle \begin{array}{l}=\frac{{\left( {10x-25} \right)}}{5}\\=\frac{{5\left( {2x-5} \right)}}{5}\end{array}\)
= (2x − 5)
(ii) (10x – 25) ÷ (2x – 5)
\(\displaystyle \begin{array}{l}=\frac{{\left( {10x-25} \right)}}{{\left( {2x-5} \right)}}\\=\frac{{5\left( {2x-5} \right)}}{{\left( {2x-5} \right)}}\end{array}\)
\(\displaystyle \begin{array}{l}=\frac{{\left( {10x-25} \right)}}{{\left( {2x-5} \right)}}\\=\frac{{5\left( {2x-5} \right)}}{{\left( {2x-5} \right)}}\end{array}\)
= 5
(iii) 10y (6y + 21) ÷ 5 (2y + 7)
\(\displaystyle \begin{array}{l}=\frac{{10y\left( {6y+21} \right)}}{{5\left( {2y+7} \right)}}\\=\frac{{10y\times 3\left( {2y+7} \right)}}{{5\left( {2y+7} \right)}}\end{array}\)
\(\displaystyle \begin{array}{l}=\frac{{10y\left( {6y+21} \right)}}{{5\left( {2y+7} \right)}}\\=\frac{{10y\times 3\left( {2y+7} \right)}}{{5\left( {2y+7} \right)}}\end{array}\)
= 2y × 3
= 6y
(iv) 9x²y² (3z -24) ÷ 27xy (z – 8)
\(\displaystyle \begin{array}{l}=\frac{{9{{x}^{2}}{{y}^{2}}\left( {3z-24} \right)}}{{27xy\left( {z-8} \right)}}\\=\frac{{9{{x}^{2}}{{y}^{2}}\times 3\left( {z-8} \right)}}{{27xy\left( {z-8} \right)}}\\=\frac{{27{{x}^{2}}{{y}^{2}}\left( {z-8} \right)}}{{27xy\left( {z-8} \right)}}\end{array}\)
\(\displaystyle \begin{array}{l}=\frac{{9{{x}^{2}}{{y}^{2}}\left( {3z-24} \right)}}{{27xy\left( {z-8} \right)}}\\=\frac{{9{{x}^{2}}{{y}^{2}}\times 3\left( {z-8} \right)}}{{27xy\left( {z-8} \right)}}\\=\frac{{27{{x}^{2}}{{y}^{2}}\left( {z-8} \right)}}{{27xy\left( {z-8} \right)}}\end{array}\)
= xy
(v) 96abc (3a – 12) (5b – 30) ÷ 144 (a – 4) (b – 6)
\(\displaystyle \begin{array}{l}=\frac{{96abc\left( {3a-12} \right)\left( {5b-30} \right)}}{{144\left( {a-4} \right)\left( {b-6} \right)}}\\=\frac{{96abc\times 3\left( {a-4} \right)\times 5\left( {b-6} \right)}}{{144\left( {a-4} \right)\left( {b-6} \right)}}\\=\frac{{1440abc\left( {a-4} \right)\left( {b-6} \right)}}{{144\left( {a-4} \right)\left( {b-6} \right)}}\end{array}\)
\(\displaystyle \begin{array}{l}=\frac{{96abc\left( {3a-12} \right)\left( {5b-30} \right)}}{{144\left( {a-4} \right)\left( {b-6} \right)}}\\=\frac{{96abc\times 3\left( {a-4} \right)\times 5\left( {b-6} \right)}}{{144\left( {a-4} \right)\left( {b-6} \right)}}\\=\frac{{1440abc\left( {a-4} \right)\left( {b-6} \right)}}{{144\left( {a-4} \right)\left( {b-6} \right)}}\end{array}\)
= 2 × 5 × abc
= 10abc
4. निर्देशानुसार भाग दीजिए :
(i) 5 (2x + 1) (3x + 5) ÷ (2x + 1) (ii) 26xy (x + 5) (y – 4) ÷ 13x (y – 4)
(iii) 52pqr (p + q) (q + r) (r + p) ÷ 104pq (q + r) (r + p)
(iv) 20 (y + 4) (y² + 5y + 3) ÷ 5(y + 4)
(v) x (x + 1) (x + 2) (x + 3) ÷ x (x + 1)
हल :
(i) 5 (2x + 1) (3x + 5) ÷ (2x + 1)
\(\displaystyle =\frac{{5\left( {2x+1} \right)\left( {3x+5} \right)}}{{\left( {2x+1} \right)}}\)
\(\displaystyle =\frac{{5\left( {2x+1} \right)\left( {3x+5} \right)}}{{\left( {2x+1} \right)}}\)
= 5 (3x + 5)
(ii) 26xy (x + 5) (y – 4) ÷ 13x (y – 4)
\(\displaystyle =\frac{{26xy\left( {x+5} \right)\left( {y-4} \right)}}{{13x\left( {y-4} \right)}}\)
\(\displaystyle =\frac{{26xy\left( {x+5} \right)\left( {y-4} \right)}}{{13x\left( {y-4} \right)}}\)
= 2y (x + 5)
(iii) 52pqr (p + q) (q + r) (r + p) ÷ 104pq (q + r) (r + p)
\(\displaystyle \begin{array}{l}=\frac{{52pqr\left( {p+q} \right)\left( {q+r} \right)\left( {r+p} \right)}}{{104pq\left( {q+r} \right)\left( {r+p} \right)}}\\=\frac{{r\left( {p+q} \right)}}{2}\\=\frac{1}{2}r\left( {p+q} \right)\end{array}\)
\(\displaystyle \begin{array}{l}=\frac{{52pqr\left( {p+q} \right)\left( {q+r} \right)\left( {r+p} \right)}}{{104pq\left( {q+r} \right)\left( {r+p} \right)}}\\=\frac{{r\left( {p+q} \right)}}{2}\\=\frac{1}{2}r\left( {p+q} \right)\end{array}\)
(iv) 20 (y + 4) (y² + 5y + 3) ÷ 5 (y + 4)
\(\displaystyle =\frac{{20\left( {y+4} \right)\left( {{{y}^{2}}+5y+3} \right)}}{{5\left( {y+4} \right)}}\)
\(\displaystyle =\frac{{20\left( {y+4} \right)\left( {{{y}^{2}}+5y+3} \right)}}{{5\left( {y+4} \right)}}\)
= 4 (y² + 5y + 3)
(v) x (x + 1) (x + 2) (x + 3) ÷ x (x + 1)
\(\displaystyle =\frac{{x\left( {x+1} \right)\left( {x+2} \right)\left( {x+3} \right)}}{{x\left( {x+1} \right)}}\)
\(\displaystyle =\frac{{x\left( {x+1} \right)\left( {x+2} \right)\left( {x+3} \right)}}{{x\left( {x+1} \right)}}\)
= (x + 2) (x + 3)
5. व्यंजकों के गुणनखंड कीजिए और निर्देशानुसार भाग दीजिए :
(i) (y² + 7y + 10) ÷ (y + 5) (ii) (m² – 14m – 32) ÷ (m + 2)
(iii) (5p² – 25p + 20) ÷ (p – 1) (iv) 4yz (z² + 6z – 16) ÷ 2y (z + 8)
(v) 5pq (p² – q²) ÷ 2p (p + q)
(vi) 12xy (9x² – 16y²) ÷ 4xy (3x + 4y)
(vii) 39y³ (50y² – 98) ÷ 25y² (5y + 7)
हल :
(i) (y² + 7y + 10) ÷ (y + 5)
\(\displaystyle \begin{array}{l}=\frac{{\left( {{{y}^{2}}+7y+10} \right)}}{{\left( {y+5} \right)}}\\=\frac{{\left( {{{y}^{2}}+5y+2y+10} \right)}}{{\left( {y+5} \right)}}\\=\frac{{y\left( {y+5} \right)+2\left( {y+5} \right)}}{{\left( {y+5} \right)}}\\=\frac{{\left( {y+5} \right)\left( {y+2} \right)}}{{\left( {y+5} \right)}}\end{array}\)
\(\displaystyle \begin{array}{l}=\frac{{\left( {{{y}^{2}}+7y+10} \right)}}{{\left( {y+5} \right)}}\\=\frac{{\left( {{{y}^{2}}+5y+2y+10} \right)}}{{\left( {y+5} \right)}}\\=\frac{{y\left( {y+5} \right)+2\left( {y+5} \right)}}{{\left( {y+5} \right)}}\\=\frac{{\left( {y+5} \right)\left( {y+2} \right)}}{{\left( {y+5} \right)}}\end{array}\)
= (y + 2)
(ii) (m² – 14m – 32) ÷ (m + 2)
\(\displaystyle \begin{array}{l}=\frac{{\left( {{{m}^{2}}-14m-32} \right)}}{{\left( {m+2} \right)}}\\=\frac{{\left( {{{m}^{2}}-16m+2m-32} \right)}}{{\left( {m+2} \right)}}\\=\frac{{\left[ {m\left( {m-16} \right)+2\left( {m-16} \right)} \right]}}{{\left( {m+2} \right)}}\\=\frac{{\left( {m-16} \right)\left( {m+2} \right)}}{{\left( {m+2} \right)}}\end{array}\)
\(\displaystyle \begin{array}{l}=\frac{{\left( {{{m}^{2}}-14m-32} \right)}}{{\left( {m+2} \right)}}\\=\frac{{\left( {{{m}^{2}}-16m+2m-32} \right)}}{{\left( {m+2} \right)}}\\=\frac{{\left[ {m\left( {m-16} \right)+2\left( {m-16} \right)} \right]}}{{\left( {m+2} \right)}}\\=\frac{{\left( {m-16} \right)\left( {m+2} \right)}}{{\left( {m+2} \right)}}\end{array}\)
= (m – 16)
(iii) (5p² – 25p + 20) ÷ (p – 1)
\(\displaystyle \begin{array}{l}=\frac{{\left( {5{{p}^{2}}-25p+20} \right)}}{{\left( {p-1} \right)}}\\=\frac{{\left( {5{{p}^{2}}-20p-5p+20} \right)}}{{\left( {p-1} \right)}}\\=\frac{{\left[ {5p\left( {p-4} \right)-5\left( {p-4} \right)} \right]}}{{\left( {p-1} \right)}}\\=\frac{{\left( {p-4} \right)\left( {5p-5} \right)}}{{\left( {p-1} \right)}}\\=\frac{{\left( {p-4} \right)5\left( {p-1} \right)}}{{\left( {p-1} \right)}}\end{array}\)
\(\displaystyle \begin{array}{l}=\frac{{\left( {5{{p}^{2}}-25p+20} \right)}}{{\left( {p-1} \right)}}\\=\frac{{\left( {5{{p}^{2}}-20p-5p+20} \right)}}{{\left( {p-1} \right)}}\\=\frac{{\left[ {5p\left( {p-4} \right)-5\left( {p-4} \right)} \right]}}{{\left( {p-1} \right)}}\\=\frac{{\left( {p-4} \right)\left( {5p-5} \right)}}{{\left( {p-1} \right)}}\\=\frac{{\left( {p-4} \right)5\left( {p-1} \right)}}{{\left( {p-1} \right)}}\end{array}\)
= 5 (p – 4)
(iv) 4yz (z² + 6z – 16) ÷ 2y (z + 8)
\(\displaystyle \begin{array}{l}=\frac{{4yz\left( {{{z}^{2}}+6z-16} \right)}}{{2y\left( {z+8} \right)}}\\=\frac{{4yz\left( {{{z}^{2}}+8z-2z-16} \right)}}{{2y\left( {z+8} \right)}}\\=\frac{{4yz\left[ {z\left( {z+8} \right)-2\left( {z+8} \right)} \right]}}{{2y\left( {z+8} \right)}}\\=\frac{{4yz\left( {z+8} \right)\left( {z-1} \right)}}{{2y\left( {z+8} \right)}}\end{array}\)
\(\displaystyle \begin{array}{l}=\frac{{4yz\left( {{{z}^{2}}+6z-16} \right)}}{{2y\left( {z+8} \right)}}\\=\frac{{4yz\left( {{{z}^{2}}+8z-2z-16} \right)}}{{2y\left( {z+8} \right)}}\\=\frac{{4yz\left[ {z\left( {z+8} \right)-2\left( {z+8} \right)} \right]}}{{2y\left( {z+8} \right)}}\\=\frac{{4yz\left( {z+8} \right)\left( {z-1} \right)}}{{2y\left( {z+8} \right)}}\end{array}\)
= 2z (z + 8)
(v) 5pq (p² – q²) ÷ 2p (p + q)
\(\displaystyle \begin{array}{l}=\frac{{5pq\left( {{{p}^{2}}-{{q}^{2}}} \right)}}{{2p\left( {p+q} \right)}}\\=\frac{{5pq\left( {p+q} \right)\left( {p-q} \right)}}{{2p\left( {p+q} \right)}}\\=\frac{{5q\left( {p-q} \right)}}{2}\\=\frac{5}{2}q\left( {p-q} \right)\end{array}\)
\(\displaystyle \begin{array}{l}=\frac{{5pq\left( {{{p}^{2}}-{{q}^{2}}} \right)}}{{2p\left( {p+q} \right)}}\\=\frac{{5pq\left( {p+q} \right)\left( {p-q} \right)}}{{2p\left( {p+q} \right)}}\\=\frac{{5q\left( {p-q} \right)}}{2}\\=\frac{5}{2}q\left( {p-q} \right)\end{array}\)
(vi) 12xy (9x² – 16y²) ÷ 4xy (3x + 4y)
\(\displaystyle \begin{array}{l}=\frac{{12xy\left( {9{{x}^{2}}-16{{y}^{2}}} \right)}}{{4xy\left( {3x+4y} \right)}}\\=\frac{{12xy\left[ {{{{\left( {3x} \right)}}^{2}}-{{{\left( {4y} \right)}}^{2}}} \right]}}{{4xy\left( {3x+4y} \right)}}\\=\frac{{12xy\left( {3x+4y} \right)\left( {3x-4y} \right)}}{{4xy\left( {3x+4y} \right)}}\end{array}\)
\(\displaystyle \begin{array}{l}=\frac{{12xy\left( {9{{x}^{2}}-16{{y}^{2}}} \right)}}{{4xy\left( {3x+4y} \right)}}\\=\frac{{12xy\left[ {{{{\left( {3x} \right)}}^{2}}-{{{\left( {4y} \right)}}^{2}}} \right]}}{{4xy\left( {3x+4y} \right)}}\\=\frac{{12xy\left( {3x+4y} \right)\left( {3x-4y} \right)}}{{4xy\left( {3x+4y} \right)}}\end{array}\)
= 3 (3x – 4y)
(vii) 39y³ (50y² – 98) ÷ 26y² (5y + 7)
\(\displaystyle \begin{array}{l}=\frac{{39{{y}^{3}}\left( {50{{y}^{2}}-98} \right)}}{{26{{y}^{2}}\left( {5y+7} \right)}}\\=\frac{{39{{y}^{3}}\times 2\left( {25{{y}^{2}}-49} \right)}}{{26{{y}^{2}}\left( {5y+7} \right)}}\\=\frac{{39{{y}^{3}}\times 2\left[ {{{{\left( {5y} \right)}}^{2}}-{{{\left( 7 \right)}}^{2}}} \right]}}{{26{{y}^{2}}\left( {5y+7} \right)}}\end{array}\)
\(\displaystyle \begin{array}{l}=\frac{{39{{y}^{3}}\left( {50{{y}^{2}}-98} \right)}}{{26{{y}^{2}}\left( {5y+7} \right)}}\\=\frac{{39{{y}^{3}}\times 2\left( {25{{y}^{2}}-49} \right)}}{{26{{y}^{2}}\left( {5y+7} \right)}}\\=\frac{{39{{y}^{3}}\times 2\left[ {{{{\left( {5y} \right)}}^{2}}-{{{\left( 7 \right)}}^{2}}} \right]}}{{26{{y}^{2}}\left( {5y+7} \right)}}\end{array}\)
= 3y (5y – 7)
NCERT SOLUTIONS FOR CLASS 8 MATHS CHAPTER 8
ncert solutions for class 8 maths class 8th maths
Ex. 14.4
प्रश्नावली 14.4
निम्नलिखित गणितीय कथनों में त्रुटि ज्ञात करके उसे सही कीजिए :
1. 4 (x – 5) = 4x – 5 2. x (3x + 2) = 3x² + 2
3. 2x + 3y = 5xy 4. x + 2x + 3x = 5x
5. 5y + 2y + y – 7y = 0 6. 3x + 2x = 5x²
7. (2x)² + 4 (2x) + 7 = 2x² + 8x + 7
8. (2x)² + 5x = 4x + 5x = 9x
9. (3x + 2)² = 3x² + 6x +4
हल :
1. 4 (x – 5) = 4x – 5
बायां पक्ष (L.H.S) = 4 (x – 5)
= 4x – 20
अतः दायां पक्ष (R.H.S) = 4x – 20 होगा।
अतः समीकरण 4 (x – 5) = 4x – 5 होगा।
2. x (3x + 2) = 3x² + 2
बायां पक्ष (L.H.S) = x (3x + 2)
= 3x² + 2x
अतः दायां पक्ष (R.H.S) = 3x² + 2x होगा।
अतः समीकरण x (3x + 2) = 3x² + 2x होगा।
3. 2x + 3y = 5xy
बायां पक्ष (L.H.S) = 2x + 3y
अतः दायां पक्ष (R.H.S) = 2x + 3y होगा।
समीकरण 2x + 3y = 2x + 3y होगा।
4. x + 2x + 3x = 5x
बायां पक्ष (L.H.S) = x + 2x + 3x
= 6x
अतः दायां पक्ष (R.H.S) = 6x होगा।
अतः समीकरण x + 2x + 3x = 6x होगा।
5. 5y + 2y + y – 7y = 0
बायां पक्ष (L.H.S) = 5y + 2y + y – 7y
= 8y – 7y
= y
अतः दायां पक्ष (R.H.S) = y होगा।
अतः समीकरण 5y + 2y + y – 7y = y होगा।
6. 3x + 2x = 5x²
बायां पक्ष (L.H.S) = 3x + 2x
= 5x
अतः दायां पक्ष (R.H.S) = 5x होगा।
अतः समीकरण 3x + 2x = 5x होगा।
7. (2x)² + 4 (2x) + 7 = 2x² + 8x + 7
बायां पक्ष (L.H.S) = (2x)² + 4 (2x) + 7
= 4x² + 8x + 7
अतः दायां पक्ष (R.H.S) = 4x² + 8x + 7 होगा।
अतः समीकरण (2x)² + 4 (2x) + 7 = 4x² + 8x + 7 होगा।
8. (2x)² + 5x = 4x + 5x = 9x
बायां पक्ष (L.H.S) = (2x)² + 5x
= 4x² + 5x
अतः दायां पक्ष (R.H.S) = 4x² + 5x होगा।
अतः समीकरण (2x)² + 5x = 4x² + 5x होगा।
9. (3x + 2)² = 3x² + 6x +4
बायां पक्ष (L.H.S) = (3x + 2)²
= (3x)² + 2 × 3x × 2 + (2)²
= 9x² + 12x + 4
अतः दायां पक्ष (R.H.S) = 9x² + 12x + 4 होगा।
अतः समीकरण (3x + 2)² = 9x² + 12x + 4 होगा।
10. x = – 3 प्रतिस्थापित करने पर प्राप्त होता है।
(a) x² + 5x + 4 से (- 3)² + 5 (- 3) + 4 = 9 + 2 + 4 = 15 प्राप्त होता है।
हल : x = – 3 समीकरण x² + 5x + 4 में रखने पर –
= (- 3)² + 5 × (- 3) + 4
= 9 – 15 + 4
= – 15 + 13
= – 2
अतः समीकरण x² + 5x + 4 में x = – 3 रखने पर – 2 प्राप्त होगा।
(b) x² – 5x + 4 से (- 3)² – 5 (- 3) + 4 = 9 – 15 + 4 = – 2 प्राप्त होता है।
हल : x = – 3 समीकरण x² – 5x + 4 में रखने पर –
= (- 3)² – 5 × (- 3) + 4
= 9 + 15 + 4
= 28
अतः समीकरण x² – 5x + 4 में x = – 3 रखने पर 28 प्राप्त होगा।
(c) x² + 5x से (- 3)² + 5 (- 3) = – 9 – 15 = – 24 प्राप्त होता है।
हल : x = – 3 समीकरण x² + 5x में रखने पर –
= (- 3)² + 5 × (- 3)
= 9 – 15
= – 6
अतः समीकरण x² + 5x में x = – 3 रखने पर – 6 प्राप्त होगा।
11. (y – 3)² = y² – 9
हल : बायां पक्ष (L.H.S) = (y – 3)²
= y² – 2 × y × 3 + (3)²
= y² – 6y + 9
अतः दायां पक्ष (R.H.S) = y² – 6y + 9 होगा।
अतः समीकरण (y – 3)² = y² – 6y + 9 होगा।
12. (x + 5)² = z² + 25
हल : बायां पक्ष (L.H.S) = (x + 5)²
= x² + 2 × x × 5 + (5)²
= x² + 10x + 25
अतः दायां पक्ष (R.H.S) = x² + 10x + 25 होगा।
अतः समीकरण (x + 5)² = x² + 10x + 25 होगा।
13. (2a + 3b) (a – b) = 2a² – 3b²
हल : बायां पक्ष (L.H.S) = (2a + 3b) (a – b)
= 2a (a – b) + 3b (a – b)
= 2a² – 2ab + 3ab – 3b²
= 2a² + ab – 3b²
अतः दायां पक्ष (R.H.S) = 2a² + ab – 3b² होगा।
अतः समीकरण (2a + 3b) (a – b) = 2a² + ab – 3b² होगा।
14. (a + 4) (a + 2) = a² + 6a +8
हल : बायां पक्ष (L.H.S) = (a + 4) (a + 2)
= a (a + 2) + 4 (a + 2)
= a² + 2a + 4a + 8
= a² + 6a + 8
अतः दायां पक्ष (R.H.S) = a² + 6a + 8 होगा।
अतः समीकरण (a + 4) (a + 2) = a² + 6a + 8 होगा।
15. (a – 4) (a – 2) = a² – 8
हल : बायां पक्ष (L.H.S) = (a – 4) (a – 2)
= a (a – 2) – 4 (a – 2)
= a² – 2a – 4a + 8
= a² – 6a + 8
अतः दायां पक्ष (R.H.S) = a² – 6a + 8 होगा।
अतः समीकरण (a – 4) (a – 2) = a² – 6a + 8 होगा।
16. \(\displaystyle \frac{{3{{x}^{2}}}}{{3{{x}^{2}}}}=0\)
हल : बायां पक्ष (L.H.S) = \(\displaystyle \frac{{3{{x}^{2}}}}{{3{{x}^{2}}}}\)
= 1
अतः दायां पक्ष (R.H.S) = 1 होगा।
अतः समीकरण $frac{{3{x^2}}}{{3{x^2}}} = 1$ होगा।
17. \(\displaystyle \frac{{3{{x}^{2}}+1}}{{3{{x}^{2}}}}=1+1=2\)
हल : बायां पक्ष (L.H.S) = \(\displaystyle \frac{{3{{x}^{2}}+1}}{{3{{x}^{2}}}}\)
\(\displaystyle \begin{array}{l}\frac{{3{{x}^{2}}}}{{3{{x}^{2}}}}+\frac{1}{{3{{x}^{2}}}}\\1+\frac{1}{{3{{x}^{2}}}}\end{array}\)
अतः दायां पक्ष (R.H.S) = \(\displaystyle 1+\frac{1}{{3{{x}^{2}}}}\) होगा।
अतः समीकरण \(\displaystyle \frac{{3{{x}^{2}}+1}}{{3{{x}^{2}}}}=1+\frac{1}{{3{{x}^{2}}}}\) होगा।
18. \(\displaystyle \frac{{3x}}{{3x+2}}=\frac{1}{2}\)हल : बायां पक्ष (L.H.S) = \(\displaystyle \frac{{3x}}{{3x+2}}\)
अतः दायां पक्ष (R.H.S) = \(\displaystyle \frac{{3x}}{{3x+2}}\) होगा।
अतः समीकरण \(\displaystyle \frac{{3x}}{{3x+2}}=\frac{{3x}}{{3x+2}}\) होगा।
19. \(\displaystyle \frac{3}{{4x+3}}=\frac{1}{{4x}}\)हल : बायां पक्ष (L.H.S) = \(\displaystyle \frac{3}{{4x+3}}\)
अतः दायां पक्ष (R.H.S) = \(\displaystyle \frac{3}{{4x+3}}\) होगा।
अतः समीकरण \(\displaystyle \frac{3}{{4x+3}}=\frac{3}{{4x+3}}\) होगा।
20. \(\displaystyle \frac{{4x+5}}{{4x}}=5\)
हल : बायां पक्ष (L.H.S) = \(\displaystyle \frac{{4x+5}}{{4x}}\)
\(\displaystyle \frac{{4x}}{{4x}}+\frac{5}{{4x}}\)
\(\displaystyle =1+\frac{5}{{4x}}\)
\(\displaystyle =1+\frac{5}{{4x}}\)
अतः दायां पक्ष (R.H.S) = \(\displaystyle 1+\frac{5}{{4x}}\) होगा।
अतः समीकरण \(\displaystyle \frac{{4x+5}}{{4x}}=1+\frac{5}{{4x}}\) होगा।
21. \(\displaystyle \frac{{7x+5}}{5}=7x\)हल : बायां पक्ष (L.H.S) = \(\displaystyle \frac{{7x+5}}{5}\)
= \(\displaystyle \frac{{7x}}{5}+\frac{5}{5}\)= \(\displaystyle \frac{{7x}}{5}+1\)
अतः दायां पक्ष (R.H.S) = \(\displaystyle \frac{{7x}}{5}+1\) होगा।
अतः समीकरण \(\displaystyle \frac{{7x+5}}{5}=\frac{{7x}}{5}+1\) होगा
Very good solutions