MP Board Class 8th Maths Solutions Chapter 9 बीजीय व्यंजक एवं सर्वसमिकाएँ Ex 9.5

प्रश्न 1.
निम्नलिखित गुणनफलों में से प्रत्येक को प्राप्त करने के लिए उचित सर्वसमिका का उपयोग कीजिए –

  1. (x + 3) (x + 3)
  2. (2y + 5) (2y + 5)
  3. (2a – y) (2a – y)
  4. (3a – \(\frac{1}{2}\)) (3a – \(\frac{1}{2}\))
  5. (11m – 0.4) (1.1m + 0.4)
  6. (a2 + b2)(- a2 + b2)
  7. (6x – 7) (6x + 7)
  8. (- a + c) (- a + c)
  9. (\(\frac{x}{2}\) + \(\frac{3y}{4}\)) (\(\frac{x}{2}\) + \(\frac{3y}{4}\))
  10. (7a – 9b) (7a – 9b)

हल:
1. (x + 3) (x + 3)
सर्वसमिका (a + b)2 = a2 + 2ab + b2 के उपयोग से,
(x + 3)2 = x2 + 2 x 3 × x + (3)2
= x2 + 6x + 9

2. (2y + 5) (2y + 5) = (2y + 5)2
सर्वसमिका (a + b)2 = a2 + 2ab + b2 का उपयोग करने पर,
∴ (2y + 5) = (2y)2 + 2 x (2y) x 5 + (5)2
= 4y2 + 20y + 25

3. (2a – 7) (2a – 7) = (2a – 7)2
सर्वसमिका (a – b)2 = a2 – 2ab + b2 का उपयोग करने पर,
∴ (2a – 7)2 = (2a)2 – 2 x 2a x (7) + (- 7)2
= 4a2 – 28a + 49

4. (3a – \(\frac{1}{2}\)) (3a – \(\frac{1}{2}\)) = (3a – \(\frac{1}{2}\))2
सर्वसमिका (a – b)2 = a2 – 2ab + b2 का उपयोग करने पर,
∴ (3a – \(\frac{1}{2}\))2 = (3a)2 – 2 x (3a) x (\(\frac{1}{2}\)) + (- \(\frac{1}{2}\))2
= 9a2 – 3a + \(\frac{1}{4}\)

5. (1.1m – 0.4) (1.1m + 0.4)
सर्वसमिका (a – b) (a + b) = a2 – b2 का उपयोग करने पर,
∴ (1.1m – 0.4) (11m + 0.4) = (1.1m)2 – (0.4)2
= 1.21m – 0.16

6. (a2 + b2) (- a2 + b2) = (b2 + a2) (b2 – a2)
सर्वसमिका (a + b) (a – b) = a2 – b2 का उपयोग करने पर,
∴ (b2 + a2) (b2 – a2) = (b2)2 – (a2)2
= b4 – a4

7. (6x – 7) (6x + 7)
सर्वसमिका (a – b)(a + b) = a2 – b2 का उपयोग करने पर
∴ (6x – 7) (6x + 7) = (6x)2 – (7)2
= 36x2 – 49

8. (- a + c) (- a + c) = (- a + c)2
सर्वसमिका (a – b)2 = a2 – 2ab + b2 का उपयोग करने पर,
= (- a)2 – 2(a)(c) + (c)2
= a2 – 2ac + c2

9. (\(\frac{x}{2}\) + \(\frac{3y}{4}\)) (\(\frac{x}{2}\) + \(\frac{3y}{2}\)) = (\(\frac{x}{2}\) + \(\frac{3y}{4}\))2
सर्वसमिका (a + b)2 = a2 + 2ab + b2 का उपयोग करने पर,
(\(\frac{x}{2}\) + \(\frac{3y}{4}\))2 = (\(\frac{x}{2}\))2 + \(\frac{x}{2}\) x \(\frac{3y}{4}\) + (\(\frac{3y}{4}\))2
= \(\frac { x^{ 2 } }{ 4 } \) + \(\frac{3xy}{4}\) + \(\frac { 9y^{ 2 } }{ 4 } \)

10. (7a – 9b) (7a – 9b) = (7a – 9b)2
सर्वसमिका (a – b) = a2 – 2ab+ b2 का उपयोग करने पर,
∴ (7a – 7b) = (7a)2 – 2 x 7a x 9b + (9b)2
= 49a2 – 126ab + 81b2

MP Board Solutions

प्रश्न 2.
निम्नलिखित गुणनफलों को ज्ञात करने के लिए सर्वसमिका (x + a) (x + b) = x2 + (a + b) x + ab का
उपयोग कीजिए –

  1. (x + 3) (x + 7)
  2. (4x + 5) (4x + 1)
  3. (4x – 5) (4x – 1)
  4. (4x + 5) (4x – 1)
  5. (2x + 5y) (2x + 3y)
  6. (2a2 + 9) (2a2 + 5)
  7. (xyz – 4) (xyz – 2).

हल:
1. (x + 3) (x + 7)
सर्वसमिका (x + a) (x + b) = x2 + (a + b)x + ab में
a = 3 तथा b = 7 रखने पर,
(x + 3) (x + 7) = x2 + (3 + 7)x + 3 x 7
= x2 + 10x + 21

2. (4x + 5) (4x + 1)
सर्वसमिका (x + a) (x + b) = x2 + (a + b) x + ab में
x = 4x, a = 5 तथा b = -1 रखने पर,
(4x + 5) (4x + 1) = (4x)2 + (5 + 1) 4x + 5 x 1
= 16x2 + 4x + 5

3. (4x – 5) (4x – 1)
सर्वसमिका (x + a) (x + b) = x2 + (a + b) x + ab में
x= 4x, a = -5 तथा b = 1 रखने पर,
(4x – 5) (4x – 1) = (4x)2 + (- 5 – 1)4x + (-5)(-1)
= 16x2 – 24x + 5

4. (4x + 5) (4x – 1)
सर्वसमिका (x + a) (x + b) = x2 + (a + b)x + ab में
x = 4x, a = 5 तथा b = – 1 रखने पर,
(4x + 5) (4x – 1) = (4x)2 + (5 – 1)4x + (5)(-1)
= 16x2 + 16x – 5

5. (2x + 5y) (2x +3y)
सर्वसमिका (x + a) (x + b) = x2 + (a + b)x + ab में
x = 2x, a = 5y तथा b = 3y रखने पर,
(2x + 5y) (2x + 3y)= (2x)2 + (5y + 3y) (2x) +5y x 3y
= 4x2 + 16xy + 15y2

6. (2a2 + a) (2a2 + 5)
सर्वसमिका (x + a) (x + b) = x2 + (a + b)x + ab में
x = 2a2, a = 9 तथा b = 5 रखने पर,
(2a2 + 9) (2a2 + 5) = (2a2)2 + (9 + 5) 2a2 + 9 x 5
= 4a4 + 28a2 + 45

7. (xyz – 4) (xyz – 2)
सर्वसमिका (x + a) (x + b) = x2 + (a + b)x + ab में
x = xyz, a =- 4 तथा b = – 2 रखने पर,
(xyz – 4) (xyz – 2) = (xyz) + (- 4 – 2)xyz + (- 4)(- 2)
= x2y2z2 – 6xyz +8

MP Board Solutions

प्रश्न 3.
सर्वसमिका का उपयोग करते हुए निम्नलिखित वर्गों को ज्ञात कीजिए –

  1. (b – 7)2
  2. (xy + 3z)2
  3. (6x2 – 5y
  4. (\(\frac{2}{3}\)m + \(\frac{3}{2}\)n)2
  5. (0.4p – 0.54)2
  6. (2xy + 5y)2

हल:
1. (b – 7)2 सर्वसमिका (a – b)2 = a2 – 2ab + b2 से,
(b – 7)2 = b2 – 2 x b x 7 + (7)2
= b2 – 14b + 49

2. (xy + 3z)
सर्वसमिका (a + b)2 = a2 + 2ab + b2 से,
(xy + 3z)2 = (xy) + 2 × xy x 3z + (3z)2
= xy2 + 6xyz + 9z2

3. (6x2 – 5y)2
सर्वसमिका (a – b)2 = a2 – 2ab + b2 से,
(6x2 – 5y)2 = (6x2)2 – 2 x 6x2 × 5y + (5y)2
= 36x4 – 60xy + 25y2

4. (\(\frac{2}{m}\)m – \(\frac{3}{2}\)n)2
सर्वसमिका (a + b)2 = a2 + 2ab + b2 से,
(\(\frac{2}{3}\)m + \(\frac{3}{2}\)n)2 = (\(\frac{2}{3}\)m)2 + 2 x \(\frac{2}{3}\)m x \(\frac{3}{2}\)n + (\(\frac{3}{2}\)n)2
= \(\frac{4}{9}\)m2 + 2mn + \(\frac{9}{4}\)n

5. (0.4p – 0.5q)2
सर्वसमिका (a – b)2 = a2 – 2ab + b2 से
(0.4p – 0.5q) = (0.4p)2 – 2 x 0.4p x 0.5q + (0.5q)2
= 0.16p2 – 0.4pq + 0.25q2

6. (2xy + 5y)2
सर्वसमिका (a + b)2 = a2 + 2ab + b2 से,
(2xy + 5y)2 = (2xy)2 + 2 x 2xy x 5y + (5y)2
= 4x2y2 + 20xy2 + 25y2

प्रश्न 4.
सरल कीजिए –

  1. (a2 – b2)2
  2. (2x + 5)2 – (2x – 5)2
  3. (7m – 8n)2 + (7m + 8n)2
  4. (4m +5n)2 + (5m +4n)2
  5. (2.5p – 1.5q)2 – (1.5p – 2.5q)2
  6. (ab + bc) – 2ab2c
  7. (m2 – n2 – m)2 + 2m3n2

हल:
1. (a2 – b2)2 = (a2)2 – 2 x a2 x b2 + (b2)2
= a4 – 2a2b2 + b4

2. (2x + 5)2 – (2x – 5)2
= (4x2 + 20x + 25) – (4x2 – 20x + 25)
= 4x2 + 20x + 25 – 4x2 + 20x – 25
= 40x

3. (7m – 8n)2 + (7m + 8n)2
= (49m2 – 112mn + 64n2) + (49m2 + 112mn + 64n2)
= 49m2 – 112mn + 64n2 + 49m2 + 112mn + 64n2
= 98m2 + 128n2

4. (4m + 5n)2 + (5m + 4n)2
= (16m2 + 40mn + 25n2) + (25m2 + 40mn + 16m2)
= 16m2 + 40mn + 25n2 + 25m2 + 40mn + 16n2
= 41m2 + 80mn + 41n2

5. (2.5p – 1.5q)2 – (1.5p – 2.5q)2
= (6.25p2 – 7.5pq + 2.25q2) – (2.25p2 – 7.5pq + 6.25q2)
= 6.25p2 – 7.5pq + 2.25q2 – 2.25p2 + 7.5pq – 6.25q2
= 6.25p2 – 2.25p2 + 2.25q2 – 6.25q2
= 4p2 – 4q2

6. (ab + bc)2 – 2ab2c
= ab2 + 2 x ab x bc + b2c2 – 2ab2c
= ab2 + 2ab2c + b2c2 – 2ab2c
= a2b2 + b2c2

7. (m2 – n2m2) + 2m3n2
= (m2)2 – 2 x m2 x n2m + (n2m)2 + 2m3n2
= m4 – 2m3n2 + n4m2 + 2m3n2
= m4 + n4m2

MP Board Solutions

प्रश्न 5.
दर्शाइए कि –

  1. (3x + 7)2 – 84x = (3x – 7)2
  2. (9p – 5q)2 + 180pq = (9p + 54)2
  3. (\(\frac{4}{3}\)m – \(\frac{3}{4}\)n)2 + 2mn = \(\frac{16}{9}\) m2 + \(\frac{9}{16}\)n2
  4. (4pq +3q)2 – (4pq – 3q)2 = 48pq2
  5. (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a) = 0.

हल:
1. (3x + 7)2 – 84x = (3x – 7)2
L.H.S. = (3x + 7)2 – 84x
= (9x2 + 42x + 49) – 84x
= 9x2 – 42x + 49
= (3x)2 – 2 x (3x) (7) + (7)2
= (3x – 7)2 = R.H.S.

2. (9p – 5q)2 + 180pq = (9p + 5q)2
L.H.S. = (9p – 5q)2 + 180pq
= 81p2 – 90pq + 25q2 + 180pq
= 81p2 + 90pq + 25q2
= (9p)2 + 2(9p)(5q) + (5q)2
= (9p + 5q) = R.H.S.

3. (\(\frac{4}{3}\)m – \(\frac{3}{4}\)1)2 + 2mn = \(\frac{16}{9}\) m2 + \(\frac{9}{16}\)n2
L.H.S. = (\(\frac{4}{3}\)m – \(\frac{3}{4}\))2 + 2mn
= \(\frac{16}{9}\) m2 – 2mn + \(\frac{9}{16}\)n2 + 2mm
= \(\frac{16}{9}\)m2 + \(\frac{9}{16}\)n2
= R.H.S.

4. (4pq + 3q)2 – (4pq – 3q)2 = 48pq2
L.H.S. = (4pq + 3q)2 – (4pq – 3q)2
= (16p2q2 + 24pq2 + 9q2) – (16p2q2 – 24pq2 + 9q2)
= 16p2q2 + 24pq2 + 9q2 – 16p2q2 + 24pq2 – 9q2
= 48pq2 = R.H.S.

5. (a – b)(a + b) + (b – c)(b + c) + (c – a)(c + a) = 0
L.H.S. = (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a)
= a2 – b2 + b2 – c2 + c2 – a2
= 0 = R.H.S.

प्रश्न 6.
सर्वसमिकाओं के उपयोग से निम्नलिखित मान ज्ञात कीजिए –

  1. 712
  2. 992
  3. 1022
  4. 9982
  5. 5.22
  6. 297 x 303
  7. 78 x 82
  8. 8.92
  9. 1.05 x 9.5

हल:
1. 712 = (70 + 1)2
= (70)2 + 2 x 70 x 1 + (1)2
= 4900 + 140 + 1
= 5041

2. 992 = (100 – 1)2
= (100)2 – 2 x 100 x 1 + (1)2
= 10000 – 200 + 1
= 9801

3. (102)2 = (100 + 2)2
= (100)2 + 2 x 100 x 2 + (2)2
= 10000 + 400 + 4 = 10404

4. (998)2 = (1000 – 2)2
= (1000)2 – 2 x 1000 x 2 + (2)2
= 1000000 – 4000 + 4
= 996004

5. (5.2)2 = (5 + 0.2)2
= (5)2 + 2 x 5 x 0.2 + (0.2)2
= 25 + 2 + 0.04
= 27.04

6. 297 x 303 = (300 – 3) (300 + 3)
= (300)2 – (3)2
[∴ (a – b) (a + b) = a2 – b2]
= 90000 – 9
= 89991

7. 78 x 82 = (80 – 2) (80 + 2)
= (80)2 – (2)2
= 6400 – 4
= 6396

8. (8.9)2 = (9 – 0.1)2
= (9)2 – 2 x 9 x 0.1 + (0.1)2
=81 – 1.8 + 0.01
= 81.01 – 1.8
= 79.21

9. 1.05 x 9.5 = (1 + 0.05) 9.5
= 1 x 9.5 + 0.05 x 9.5
= 9.5 + 0.475
= 9.975

प्रश्न 7.
a2 – b2 = (a + b) (a – b) का उपयोग करते हुए निम्नलिखित का मान ज्ञात कीजिए –

  1. 512 – 492
  2. (1.02)2 – (0.98)2
  3. 1532 – 1472
  4. 12.12 – 7.92

हल:
1. 512 – 492
a2 – b2 = (a + b) (a-b) का उपयोग करने पर,
512 – 4a2 = (51 + 49) (51 – 49)
= 100 x 2 = 200

2. (1.02)2 – (0.98)2
a2 – b2 = (a + b) (a – b) का उपयोग करने पर
(1.02)2 – (0.98)2 = (1.02 + 0.98) (1.02 – 0.98)
= 2.00 x 0.04 = 0.08

3. 1532 – 1472
a2 – b2 = (a + b)(a – b) का उपयोग करने पर
1532 – 1472 = (153 + 147) (153 – 147)
= 300 x 6 = 1800

4. 12.12 – 7.92
a2 – b2 = (a + b) (a – b) का उपयोग करने पर,
12.12 – 7.92 = (12.1 + 7.9) (12.1 – 7.9)
= 20 x 4.2 = 84

MP Board Solutions

प्रश्न 8.
(x + a) (x + b) = x2 + (a + b)x + ab का उपयोग करते हुए निम्नलिखित मान ज्ञात कीजिए –

  1. 103 x 104
  2. 5.1 x 5.2
  3. 103 x 98
  4. 9.7 x 9.8.

हल:
1. 103 x 104 = (100 + 3) (100 + 4)
सर्वसमिका (x + a) (x + b) = x2 + (a + b)x + ab में
x = 100, a = 3 तथा b = 4 रखने पर,
(100 + 3) (100 + 4) = (100)2 + (3 + 4) 100 + 3 x 4
= 10000 + 700 + 24 = 10724

2. 5.1 x 5.2 = (5 + 0.1) (5 + 0.2)
सर्वसमिका (x + a) (x + b) = x2 + (a + b)x + ab में,
x = 5, a = 0.1 तथा b = 0.2 रखने पर,
(5 + 0.1) (5.02) = (5)2 + (0.1 + 0.2) 5 + 0.1 x 0.2
= 25 + 1.5 + 0.02 = 26.52

3. 103 x 98 = (100 + 3) (100 – 2)
सर्वसमिका (x + a) (x + b) = x2 + (a + b) x + ab में
x = 100, a = 3 तथा b = – 2 रखने पर,
(100 + 3) (100 – 2)= (100)2 + (3 – 2) 100 – 3 x (-2)
= 10000 + 100 – 6 = 10094

4. 9.7 x 9.8 = (9 + 0.7) (9 + 0.8)
सर्वसमिका (x + a) (x + b) = x2 + (a + b) x + ab में
x = 9, a = 0.7 तथा b = 0.8 रखने पर,
(9 + 0.7) (9 + 0.8) = (9)2 + (0.7+ 0.8)9 + 0.7 x 0.8
= 81 + 13.5 + 0.56 = 95.06

MP Board Class 8th Maths Solutions

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