RBSE Solutions for Class 8 Maths Chapter 9 बीजीय व्यंजक Ex 9.3

RBSE Solutions for Class 8 Maths Chapter 9 बीजीय व्यंजक Ex 9.3 is part of RBSE Solutions for Class 8 Maths. Here we have given Rajasthan Board RBSE Class 8 Maths Chapter 9 बीजीय व्यंजक Exercise 9.3.

Board RBSE
Textbook SIERT, Rajasthan
Class Class 8
Subject Maths
Chapter Chapter 9
Chapter Name बीजीय व्यंजक
Exercise Exercise 9.3
Number of Questions 8
Category RBSE Solutions

Rajasthan Board RBSE Class 8 Maths Chapter 9 बीजीय व्यंजक Ex 9.3

प्रश्न 1
उचित सर्वसमिका का उपयोग करते हुए निम्न गुणनफल ज्ञात कीजिए
(i) (x + 5) (x + 5)
(ii) (3x + 2) (3x + 2)
(iii) (5a – 7) (5a – 7)
(iv) (3p –\(\frac { 1 }{ 2 }\)) (3p – \(\frac { 1 }{ 2 }\))
(v) (1,2m – 0.3) (1.2m – 0.3)
(vi) (x2 +y2) (x2 -y2)
(vii) (6y + 7) (- 6y + 7)
(viii) (7a + 9b) (7a – 9b)
हल:
(i) (x + 5) (x + 5) = (x + 5)2
= (x)2+ 2 (x) (5) + (5)2 सर्वसमिका I से
= x + 10x + 25

(ii) (3x + 2) (3x + 2)
= (3x + 2)2
= (3x)2 + 2 (3x) (2) + (2)2 सर्वसमिका I से
= 9x2 + 12x +4

(iii) (5a – 7) (5a – 7)
= (5a – 7)2
= (5a)2 – 2 (3x) (2) + (2)2 सर्वसमिका II से
= 25a2 – 70a + 49

(iv) (3p –\(\frac { 1 }{ 2 }\)) (3p –\(\frac { 1 }{ 2 }\))
= (3p – \(\frac { 1 }{ 2 }\))2
= (3p)2 -2 (3p) x(RBSESolutions.com) (\(\frac { 1 }{ 2 }\))-(\(\frac { 1 }{ 2 }\))2 सर्वसमिका II से
= 9p2 – 3p + \(\frac { 1 }{ 4 }\)

(v) (1.2m – 0.3) (1,2m – 0.3)
= (1.2m – 0.3)2
= (1.2 m)2 – 2 (1.2 m) (0.3) + (0.3)2 सर्वसमिका II से
= 1.44 m2 – 0.72 m + 0.09

(vi) (x2 + y2) (x2 – y2)
= (x2)2 – (y2)2 सर्वसमिका II से
= x4 – y4

(vii) (6y + 7) (-6y + 7)
= (7 + 6y) (7 – 6y)
= (7)2 – (6y)2 सर्वसमिका II से
= 49 – 36y2

(viii) (7a + 9b) (7a – 9b)
(7a + 9b)(RBSESolutions.com) (7a – 9b) = (7a)2 – (9b)2 सर्वसमिका III से
= 49a2 – 81b2

RBSE Solutions

प्रश्न 2
निम्न व्यंजकों का गुणा सर्वसमिका (x + a) (x + b) = x2 + (a + b) x + ab का उपयोग करते हुए कीजिए
(i) (x + 1) (x + 2)
(ii) (3x + 5) (3 + 1)
(iii) (4r – 5) (4 – 1)
(iv) (30 + 5) (30 – 8)
(v) (xyz – 1) (xyz – 2)
हल:
(i) (x + 1) (x + 2)
(x + 1) (x + 2) = x2+ (1 + 2) x + 1 x 2 प्रदत्त सर्वसमिका से
= x2 + 3x + 2

(ii) (3x + 5) (3x + 1)
(3x + 5) (3x + 1) = (3x)2 + (5 + 1) 3x +5 x 1 प्रदत्त सर्वसमिका से
= 9x2 + 18 + 5

(iii) (4x – 5) (4x – 1)
(4x – 5) (4x – 1) = {4x + (-5)} {4x + (-1)}
= (4x)2 + {(- 5) + (- 1)} 4x + (-5) (-1) प्रदत्त सर्वसमिका से
= 16x2 + (-6) 4x + (5)
= 16x2 – 24x + 5

(iv) (3a + 5) (3a – 8)
(3a + 5) (34 – 8)
= (3a + 5) {3a + (- 8)} = (3a)2 + {5 + (- 8)} 34 + (5) (- 8) प्रदत्त सर्वसमिका से
= 9a2 + (- 3) 3a – 40
= 9a2 – 9a – 40

(v) (xyz – 1) (xyz – 2)
(xyz – 1) (zyz – 2) = {xyz + (- 1)} {xyz + (-2)}
= (xyz)2 + {(- 1) + (-2)} xyz + (-1) (-2)
= x2y2z2 – 3xyz + 2

RBSE Solutions

प्रश्न 3
सर्वसमिका का उपयोग करते हुए(RBSESolutions.com) निम्नलिखित वर्गों को ज्ञात कीजिए
(i) (b – 7)2
(i) (xy + 3z)2
(ii) (6m2 – 5n)2
(iv) ( \(\frac { 3 }{ 2 }\)x + \(\frac { 2 }{ 3 }\)y)2
हल:
(i) (b – 7)2 = (b)2 – 2 (b) (7) + (7)2 सर्वसमिका II से
= b2 – 14b + 49

(ii) (xy + 3z)2 = (xy)2 + 2 (xy) (3z) + (3z)2 सर्वसमिका I से
= x2y2 + 6xyz + 9z2

(iii) (6m2 – 5n)2
= (6m2)2 – 2(6m)2 (5n) (5n)2 सर्वसमिका II से
= 36m4 – 60m2n + 25n2
RBSE Solutions for Class 8 Maths Chapter 9 बीजीय व्यंजक Ex 9.3 Q1

RBSE Solutions

प्रश्न 4
सरल कीजिए
(i) (a2 – b2)
(ii) (2n + 5)2 – (2n – 5)2
(iii) (7m – 8n)2 + (7m + 8n)2
(iv) (m2 – n2m)2 + 2m
हल:
(i) (a2 – b2)2
= (a2)2 – 2(a2) (b2) + (b2)2 सर्वसमिका II से
= a4 – 2a2b2 + b4

(ii) (2n + 5)2 – (2n – 5)2
(2n + 5)2 – (2n – 5)2 = {(2n)2 + 2 (2n) (5) + (5)2} – {(2n)2 – 2(2n) (5) + (5)2} सर्वसमिका I व सर्वसमिका II से
= (4n2 + 20n + 25) – (4n2 – 20n + 25)
= 4n2 + 20n + 25 – 4n2 + 20n – 25
= 4n2 – 4n2 + 20n + 20n + 25 – 25
= 0 + 40n + 0
= 40n
वैकल्पिक विधि
(2n + 5)2– (2n – 5)2 = {(2n + 5) + (2n -5)} {(2n + 5) – (2n – 5)} सर्वसमिका II से
= (4n) (10)
= 40n

(iii) (7m – 8n)2 + (7m + 8n)2 = {(7m)2 – 2(7m) (8n) + (8n)2} + {(7m)2 + 2(7m) (8n) + (8n)2} सर्वसमिका II और सर्वसमिका I से
= (49m2 – 112mn +64n2) + (49m2 + 112mn + 64n2)
= 49m2 – 112mn + 64n2 + 49m2 + 112mn + 64n2
= 49m2+ 49m2 – 112mn + 112mn + 64m2 + 64n2
= 98m2 + 128n2

(iv) (m2 -n2m2) + 2m2n2 = [(m2) 2– 2(m2) (n2m) + (n2m2)] + 2m2n2 सर्वसमिका II से
= m4 – 2m3n2 + n4m2 + 2m3n2
= m4 – 2m3n2 + 2m3n2 + n4m2
= m4 + n4m2

RBSE Solutions

प्रश्न 5
दर्शाइए कि
(i) (2a + 3b)2 – (2a – 3b)2 = 24ab
(ii) (4x + 5)2 – 80x = (4x – 5)2
(iii) (3x – 2y)2 + 24xy = (3x + 2y)2
(iv) (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a) = 0
हल:
(i) LHS
= (2a + 3b)2 – (2a – 3b)2
= [(2a)2 + 2(2a) (3a) + (3b)2] – [(2a)2 – 2(2a) (3a) + (3b)2] सर्वसमिका I व सर्वसमिका II से
= (4a2 + 12ab + 9b2) – (4a2 – 12ab + 9b2)
= 4a+ 12ab + 9b2 – 4a2 + 12ab – 9b2
= (4a2 – 4a2 )(RBSESolutions.com) + (12ab + 12ab) + (9b2 – 9b2)
= 0 + 24ab + 0
= 24ab = RHS

(ii) LHS
(4x + 5)2 – 80x
= [(4x)2 + 2(47) (5) + (5)2] – 80x सर्वसमिका II से
= (16x2 + 40x + 25) – 80x
= 16x2 + 40x – 80x + 25
= 16x2 – 40x + 25
= (4x)2 – 2(4x) (5) + (5)2
= (4x – 5)2 सर्वसमिका II से
= RHS

(iii) LHS
(3x – 2y)2 + 24xy
= [(3x)2 – 2(3x) (2y) + (2y)2] + 24xy सर्वसमिका II से
= (9x2 – 12xy + 4y2) + 24xy
= 9x2 – 12xy + 24xy + 4y2
= 9x2 + 12xy + 4y2
= (3x)2 + 2 (3x) (2y) + (2y)2
= (3x + 2y)2 सर्वसमिका I से
= RHS

(iv) LHS
(a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a)
= (a2 – b2) + (b2 – c2) + (c2– a2)
= a2 – b2 + b2 – c2 + c2 – a2
= (a2 – a2) + (b2 – b2) + (c2 – c2)
= 0 + 0 + 0 + 0
= RHS

RBSE Solutions

प्रश्न 6
उपयुक्त सर्वसमिका का उपयोग करते हुए निम्नलिखित का मान ज्ञात कीजिए
(i) 992
(ii) 1032
(iii) 297 x 303
(iv) 78 x 82
हल:
(i) 992
992 = (100 – 1)2
= (100)2 – 2 (100) (1) + (1)2 सर्वसमिका II से
= 10000 – 200 + 1
= 10000 + 1 – 200
= 10001 – 200
= 9801

(ii) 1032
1032 = (10) + 3)2 = (100)2 + 2 (100) (3) + (3)2 सर्वसमिका I से
= 10000 + 600 +9
= 10609

(iii) 297 x 303
297 x 303 = (300 – 3) x (300 + 3)
= (300)2 – (3)2 सर्वसमिका III से
= 90000 – 9
= 89991

(iv) 78 x 82
78 x 82 = (80 – 2) x (80 + 2)
= (80)2 – (2)2 सर्वसमिका(RBSESolutions.com) III से
= 6400 – 4
= 6396

RBSE Solutions

प्रश्न 7
a2 – b2 = (a + b) (a-b) का उपयोग करते हुए निम्नलिखित का मान ज्ञात कीजिए
(i) 1012 – 992
(ii) (10.3)2 – (9.7)2
(iii) 1532 – 1472
हल:
(i) 1012 – 9922
1012 – 992 = (101 + 99) (101 – 99) प्रदत्त सर्वसमिका से
= (200) (2)
= 400

(ii) (10.3)2 – (9.7)2
(10.3)2 – (9.7)2 = (10.3 + 9.7) (10.3 – 9.7) प्रदत्त(RBSESolutions.com) सर्वसमिका से
= (20) (0.6)
= 12

(iii) 1532 – 1472
1532 – 1472 = (153 + 147) (153 – 147) प्रदत्त सर्वसमिका से
= (300) (6)
= 1800

RBSE Solutions

प्रश्न 8
(x + a) (x + b) = x2 + (a + b)x + ab का उपयोग(RBSESolutions.com) करते हुए निम्नलिखित का मान ज्ञात कीजिए
(i) 103 x 102
(ii) 7.1 x 7.3
(ii) 102 x 99
(iv) 9.8 x 9.6
हल:
(i) 103 x 102
103 x 102 = (100 + 3) x (10) + 2)
= (100)2 + (3 + 2) 100 + 3 x 2 प्रदत्त सर्वसमिका से
= 10000 + 500 + 6
= 10506

(ii) 7.1 x 7.3
7.1 x 7.3 = (7 + 0.1) x (7 + 0.3)
= (7)2+ (0.1 + 0.3) 7 + (0.1) x (0.3) प्रदत्त(RBSESolutions.com)सर्वसमिका से
= 49 + 2.8 + 0.03 = 51.83

(iii) 102 x 99
102 x 99 = (100 + 2) x (100 – 1)
= (100 + 2) X {100 + (- 1)}
= (100)2 + {2 + (- 1)} 100 + (2) (-1) प्रदत्त सर्वसमिका से
= 10000 + 100 – 2
= 10098

(iv) 9.8 x 9.6
9.8 x 9.6 = (10 – 0.2) x (10 – (0.4)
= {10 + (- 0.2)} x {10 + (-0.4)}
= (10)2 + {(- 0.2) + (-0.4)} 10 + (-0.2) (- 0.4) प्रदत्त सर्वसमिका से
= 100 – 6 + 0.08
= 94,08

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