NCERT Solutions Class 8th Math Chapter – 9 बीजीय व्यंजक एवं सर्वसमिकाएँ प्रश्नावली 9.5

May 17, 2022

NCERT Solutions Class 8th Math Chapter – 9 बीजीय व्यंजक एवं सर्वसमिकाएँ

Textbook	NCERT
Class	8^th
Subject	(गणित) Mathematics
Chapter	9^th
Chapter Name	बीजीय व्यंजक एवं सर्वसमिकाएँ
Mathematics	Class 8th गणित
Medium	Hindi
Source	Last Doubt

NCERT Solutions Class 8th Math Chapter – 9 बीजीय व्यंजक एवं सर्वसमिकाएँ

Chapter – 9

बीजीय व्यंजक एवं सर्वसमिकाएँ

प्रश्नावली 9.5

Ncert Solution Class 8th (Chapter – 9) Exercise – 9.5 Question . 1

1. निम्नलिखित गुणनफलो में से प्रत्येक को प्राप्त करने के लिए उचित सर्वसमिका का उपयोग कीजिए :

(i) (x + 3) (x + 3)

हल:
(i) (x + 3) (x + 3)
= (x + 3)² {(a + b)² = a² + b² + 2ab सर्वसमिकाएँ का उपयोग}
= x² + 6x + 9

(ii) (2y + 5) (2y + 5)

हल:
(ii) (2y + 5) (2y + 5)
= (2y + 5)²{(a + b)² = a² + b² + 2ab सर्वसमिकाएँ का उपयोग}
= 4y² + 20y + 25

(iii) (2a – 7) (2a – 7)

हल:
(iii) (2a – 7) (2a – 7)
= (2a – 7)²{(a – b)² = a² + b² – 2ab सर्वसमिकाएँ का उपयोग}
= 4a² – 28a + 49

(iv) (3a – 1/2) (3a – 1/2)

हल:
(iv) (3a – 1/2) (3a – 1/2)
= (3a – 1/2)²{(a – b)² = a² + b² – 2ab सर्वसमिकाएँ का उपयोग}
= 9a² – 3a + (1/4)

(v) (1.1m – 0.4) (1.1m + 0.4)

हल:
(v) (1.1m – 0.4) (1.1m + 0.4) {(a – b)(a + b) = a² – b² सर्वसमिकाएँ का उपयोग}
= 1.21m² – 0.16

(vi) (a²+ b²) (- a²+ b²)

हल:
(vi) (a² + b²) (- a² + b²) {(a – b) (a + b) = a² – b² सर्वसमिकाएँ का उपयोग}
= (b² + a²) (b² – a²)
= b⁴– a⁴

(vii) (6x – 7) (6x + 7)

हल:
(vii) (6x – 7) (6x + 7) {(a – b)(a + b) = a² – b² सर्वसमिकाएँ का उपयोग}
= 36x² – 49

(viii) (- a + c) (- a + c)

हल:
(viii) (- a + c) (- a + c)
= (- a + c)²{(a + b)² = a² + b² – 2ab सर्वसमिकाएँ का उपयोग}
= c² + a² – 2ac

(ix) (1/2x + 3/4y) (1/2x + 3/4y)

हल:
(ix) (x2/4) + (9y2/16) + (3xy/4) {(a + b)² = a² + b² + 2ab सर्वसमिकाएँ का उपयोग}

(x) (7a – 9b) (7a – 9b)

हल:
(x) (7a – 9b) (7a – 9b)
= (7a – 9b)²{(a + b)² = a² + b² – 2ab सर्वसमिकाएँ का उपयोग}
= 49a² – 126ab + 81b²

Ncert Solution Class 8th (Chapter – 9) Exercise – 9.5 Question . 2

2. निम्नलिखित गुणनफलो को ज्ञात करने के लिए, सर्वसमिका (x + a)(x + b) = x²+ (a + b)x + ab का उपयोग कीजिए :

(i) (x + 3) (x + 7)

हल:
(i) (x + 3) (x + 7)
= x² + (3 + 7)x + 21
= x² + 10x + 21

(ii) (4x + 5) (4x + 1)

हल:
(ii) (4x + 5) (4x + 1)
= 16x² + 4x + 20x + 5
= 16x² + 24x + 5

(iii) (4x – 5) (4x – 1)

हल:
(iii) (4x – 5) (4x – 1)
= 16x² – 4x – 20x + 5
= 16x² – 24x + 5

(iv) (4x + 5) (4x – 1)

हल:
(iv) (4x + 5) (4x – 1)
= 16x² + (5 – 1)4x – 5
= 16x² + 16x – 5

(v) (2x + 5y) (2x + 3y)

हल:
(v) (2x + 5y) (2x + 3y)
= 4x² + (5y + 3y)2x + 15y²
= 4x² + 16xy + 15y²

(vi) (2a² + 9) (2a² + 5)

हल:
(vi) (2a²+ 9) (2a²+ 5)
= 4a⁴ + (9 + 5)2a² + 45
= 4a⁴ + 28a² + 45

(vii) (xyz – 4) (xyz – 2)

हल:
(vii) (xyz – 4) (xyz – 2)
= x²y²z² + (- 4 – 2)xyz + 8
= x²y²z² – 6xyz + 8

Ncert Solution Class 8th (Chapter – 9) Exercise – 9.5 Question . 3

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

(i) (b – 7)²

हल:
(i) (b – 7)²{(a – b)(a + b) = a² – b² सर्वसमिकाएँ का उपयोग}
= b² – 14b + 49

(ii) (xy + 3z)²

हल:
(ii) (xy + 3z)²{(a + b)² = a² + b² + 2ab सर्वसमिकाएँ का उपयोग}
= x²y² + 6xyz + 9z²

(iii) (6x² – 5y)²

हल:
(iii) (6x² – 5y)²{(a – b)(a + b) = a² – b² सर्वसमिकाएँ का उपयोग}
= 36x⁴ – 60x²y + 25y²

(iv) [(2m/3) + (3n/2)]²

हल:
(iv) [( 2m/3}) + (3n/2)]²{(a + b)² = a² + b² + 2ab सर्वसमिकाएँ का उपयोग}
= (4m²/9) + (9n²/4) + 2mn

(v) (0.4p) ) – 0.5q)²

हल:
(v) (0.4p – 0.5q)²{(a – b)(a + b) = a² – b² सर्वसमिकाएँ का उपयोग}
= 0.16p² – 0.4pq + 0.25q²

(vi) (2xy + 5y)²

हल:
(vi) (2xy + 5y)²{(a + b)² = a² + b² + 2ab सर्वसमिकाएँ का उपयोग}
= 4x²y² + 20xy² + 25y²

Ncert Solution Class 8th (Chapter – 9) Exercise – 9.5 Question . 3

4. सरल कीजिए :

(i) (a² – b²)²

हल:
(i) (a²– b²)²
= a⁴ + b⁴ – 2a²b²

(ii) (2x + 5)² – (2x – 5)²

हल:
(ii) (2x + 5)² – (2x – 5)²
= 4x² + 20x + 25 – (4x² – 20x + 25)
= 4x² + 20x + 25 – 4x² + 20x – 25
= 40x

(iii) (7m – 8n)² + (7m + 8n)²

हल:
(iii) (7m – 8n)² + (7m + 8n)2
= 49m² – 112mn + 64n² + 49m² + 112mn + 64n²
= 98m² + 128n²

(iv) (4m + 5n)² + (5m + 4n)²

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

(v) (2.5p – 1.5q)² – (1.5p – 2.5q)²

हल:
(v) (2.5p – 1.5q)² – (1.5p – 2.5q)²
= 6.25p² – 7.5pq + 2.25q² – 2.25p² + 7.5pq – 6.25q²
= 4p²– 4q²

(vi) (ab + bc)²– 2ab²c

हल:
(vi) (ab + bc)²– 2ab²c
= a²b² + 2ab²c + b²c² – 2ab²c
= a²b² + b²c²

(vii) (m² – n²m)² + 2m3n²

हल:
(vii) (m² – n²m)² + 2m3n
2 = m⁴ – 2m3n² + m²n⁴ + 2m³n
2 = m⁴ + m²n⁴

5. दर्शाइए कि :

(i) (3x + 7)2 – 84x = (3x – 7)2
(ii) (9p – 5q)2+ 180pq = (9p + 5q)2
(iii) (4/3m – 3/4n)2 + 2mn = 16/9 m2 + 9/16 n2
(iv) (4pq + 3q)2– (4pq – 3q)2 = 48pq2
(v) (a – b) (a + b) + (b – c) (b) + c) + (c – a) (c + a) = 0

हल:

(i) LHS = (3x + 7)2 – 84x
= 9×2 + 42x + 49 – 84x
= 9×2 – 42x + 49
= RHS LHS
= RHS

(ii) LHS = (9p – 5q)2+ 180pq
= 81p2 – 90pq + 25q2 + 180pq
= 81p2 + 90pq + 25q2
RHS = (9p + 5q)2
= 81p2 + 90pq + 25q2
LHS = RHS

(iv) LHS = (4pq + 3q)2– (4pq – 3q)2
= 16p2q2 + 24pq2 + 9q2 – 16p2q2 + 24pq2 – 9q2
= 48pq2
= RHS
LHS = RHS

(b) LHS = (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a)
= A 2 – B 2 + B 2 – C 2 + C 2 – A
2 = 0
= RHS

6. सर्वसामिकाओ के उपयोग से निम्नलिखित मान ज्ञात कीजिए :

(i) 71²
(ii) 99²
(iii) 1022
(iv) 998²
(v) 5.2²
(vi) 297 x 303
(vii) 78 x 82
(viii) 8.92
(ix) 10.5 x 9.5

हल:

(i) 712
= (70+1)2
= 702 + 140 + 12
= 4900 + 140 +1
= 5041

(ii) 99²
= (100 -1)2
= 1002 – 200 + 12
= 10000 – 200 + 1
= 9801

(iii) 1022
= (100 + 2)2
= 1002 + 400 + 22
= 10000 + 400 + 4 = 10404

(iv) 9982
= (1000 – 2)2
= 10002 – 4000 + 22
= 1000000 – 4000 + 4
= 996004

(v) 5.22
= (5 + 0.2)2
= 52 + 2 + 0.22
= 25 + 2 + 0.04 = 27.04

(vi) 297 x 303
= (300 – 3) (300 + 3)
= 3002 – 32
= 90000 – 9
= 89991

(vii) 78 x 82
= (80 – 2) (80 + 2)
= 802 – 22
= 6400 – 4
= 6396

(viii) 8.92
= (9 – 0.1)2

= 92 – 1.8 + 0.12
= 81 – 1.8 + 0.01
= 79.21

(ix) 10.5 x 9.5
= (10 + 0.5) (10 – 0.5)
= 102 – 0.52

= 100 – 0.25
= 99.75

7. $a^{2} - b^{2} = (a + b) (a - b)$ का उपयोग करते हुए , निम्नलिखित मान ज्ञात कीजिए :

(i) 512-492
(ii) (1.02) 2- (0.98) 2
(iii) 1532-1472
(iv) 12.12-7.92

हल:

(i) 512-492
= (51 + 49) (51-49) = 100 x 2 = 200

(ii) (1.02)2- (0.98)2
= (1.02 + 0.98)(1.02 – 0.98) = 2 x 0.04 = 0.08

(iii) 1532 – 1472
= (153 + 147)(153 – 147) = 300 x 6 = 1800

(iv) 12.12 – 7.92
= (12.1 + 7.9) (12.1 – 7.9) = 20 x 4.2 = 84

8. $(x + a) (x + b) = x^{2} + (a + b) x + a b$ का उपयोग करते हुए , निम्नलिखित मान ज्ञात कीजिए : (i) 103 x 104
(ii) 5.1 x 5.2
(iii) 103 x 98
(iv) 9.7 x 9.8

हल:

(i) 103 x 104
= (100 + 3) (100 + 4)
= 1002 + (3 + 4)100 + 12
= 10000 + 700 + 12
= 10712

(ii) 5.1 x 5.2
= (5 + 0.1)(5 + 0.2)
= 52 + (0.1 + 0.2)5 + 0.1 x 0.2
= 25 + 1.5 + 0.02
= 26.52

(iii) 103 x 98
= (100 + 3)(100 – 2)
= 1002 + (3-2)100 – 6
= 10000 + 100 – 6
= 10094

(iv) 9.7 x 9.8
= (9 + 0.7) (9 + 0.8)
= 92 + (0.7 + 0.8)9 + 0.56
= 81 + 13.5 + 0.56
= 95।