Balaji Class 9 Maths Solutions Chapter 4 Algebraic Identities Ex 4.3 बीजगणितीय सर्वसमिकाऐं
Ex 4.3 Algebraic Identities अतिलघु उत्तरीय प्रश्न (Very Short Answer Type Questions)
प्रश्न 1.
(2x + 1)3 का मान ज्ञात कीजिए। [NCERT]
हलः
(2x + 1)3 = (2x)3 + (1)3 + 3 × 2x × 1(2x + 1)
= 8x3 + 1 + 12x2 + 6x
प्रश्न 2.
(2a – 3b) का मान ज्ञात कीजिए। [NCERT]
हलः
(2a – 3b)3 = (2a)3 + (-3b)3 + 3(2a) (-3b)(2a – 3b)
= 8a3 – 27b3 – 18ab(2a – 3b)
= 8a3 – 27b3 – 36a2b + 54ab2
Ex 4.3 Algebraic Identities लघु उत्तरीय प्रश्न – I (Short Answer Type Questions – I)
प्रश्न 3.
निम्न व्यंजकों के घन का मान ज्ञात कीजिए। (NCERT Exemplar)
(i) [latex]\left(\frac{1}{x}+\frac{y}{3}\right)[/latex]
(ii) (4 – [latex]\frac{1}{3 x}[/latex])
(iii) (3x – 2y)
हलः
प्रश्न 4.
निम्न को सरल कीजिए।
(i) (4x + 2y)3 + (4x – 2y)3
(ii) (4x + 2y)3 – (4x – 2y)3
(iii) (a + 3)3 + (a – 3)3
हलः
(i) (4x + 2y)3 = (4x)3 + (2y)3 + 3. (4x) . (2y) (4x + 2y)
= 64x3 + 8y3 + 24xy (4x + 2y)
= 64x3 + 8y3 + 96x2y + 48xy2
(4x – 2y)3 = (4x)3 + (-2y)3 + 3. (4x) (-2y) (4x -2y)
= 64x3 – 8y3 – 24xy (4x -2y) –
= 64x3 – 8y3 – 96x2y + 48xy2
इसलिए (4x + 2y)3 + (4x – 2y)3 = 128x3 +96xy2
(ii) (4x + 2y)3 – (4x – 2y)3
= 64x3 + 8y3 + 96x2y + 98xy2 – (64x3 – 8y3 – 96x2y + 48xy2)
= 64x3 + 8y3 + 96x32y + 48xy2 – 64x3 + 8y3 + 96x2y – 48xy2
= 16y3 + 192x2y
(iii) (a + 3)3 = a3 + 27 + 3a . 3(a + 3)
= a3 + 27 + 9a2 + 27a
(a – 3)3 = a3 – 27 – 3a – 3(a – 3)
= a3 – 27 – 9a2 + 27a
∴ (a + 3)3 + (a – 3)3 = 2a3 +54a
Ex 4.3 Algebraic Identities लघु उत्तरीय प्रश्न – II (Short Answer Type Questions – II)
प्रश्न 5.
यदि a – [latex]\frac{1}{a}[/latex] = 7, तब [latex]a^{3}-\frac{1}{a^{3}}[/latex] का मान ज्ञात कीजिए।
हलः
प्रश्न 6.
यदि a – b = 4 व ab = 21, तब a3 – b3 का मान ज्ञात कीजिए।
हलः
a – b = 4
वर्ग करने पर
a2 + b2 – 2ab = 16
a2 + b2 – 2 × 21 = 16
a2 + b2 = 16 + 42 = 58
अब a3 – b3 = (a – b)(a + b2 + ab)
= (4)(58 + 21)
= (4)(79) = 316
प्रश्न 7.
यदि x + y = 10 व xy = 21, तब x3 + y3 का मान ज्ञात कीजिए।
हलः
x + y = 10 ……………. (1)
घन करने पर
(x + y)3 = (10)3
x3 + y3 + 3xy (x + y) = 1000
x3 + y3 + 3. 21 (10) = 1000
x3 + y3 + 630 = 1000
x3 + y 3 = 1000 – 630 = 370
प्रश्न 8.
यदि 3x – 2y =11 व xy = 12, तब 27x3 – 8y3 का मान ज्ञात कीजिए।
हल:
∵ 3x – 2y = 11 …………….. (1)
वर्ग करने पर
9x2 + 4y2 – 12xy = 121
9x2 + 4y2 -12 × 12 = 121
9x2 + 4y2 = 121 + 144 = 265
27x3 – 8y3 = (3x)3 – (2y)3
= (3x – 2y)(9x2 + 4y2 + 6xy)
= (11)(265 + 6 × 12)
= (11)(265 + 72)
= (11) (337) = 3707
प्रश्न 9.
यदि [latex]a^{2}+\frac{1}{a^{2}}[/latex] = 98, तब [latex]a^{3}+\frac{1}{a^{3}}[/latex] का मान ज्ञात कीजिए।
हल:
प्रश्न 10.
यदि [latex]a^{2}+\frac{1}{a^{2}}[/latex] = 51, तब [latex]a^{3}-\frac{1}{a^{3}}[/latex] का मान ज्ञात कीजिए। .
हल:
Ex 4.3 Algebraic Identities दीर्घ उत्तरीय प्रश्न (Long Answer Type Questions)
प्रश्न 11.
निम्न के मान ज्ञात कीजिए।
(i) (99)3
(ii) (9.9)3
(iii) (10.4)3
(iv) (598)3
(v) (402)3
(vi) (1002)3
हलः
(i) (99)3 = (100 – 1)3
___ = (100) – (1)3 – 3 × 100 × 1(100 – 1)
= 1000000 – 1 – 30000+ 300 = 970299
(ii) (9.9)3 = (10 – 0.1)3
= (10)3 – (0.1)3 – 3 × 10 × 0.1(10 – 0.1)
= 1000 – 0.001 – 3(9.9)
= 1000 – 0.001 – 29.7 = 970.299
(iii) (10.4)3 = (10 + 0.4)3
= (10)3 + (0.4)3 + 3 × 10 × 0.4(10 + 0.4)
= 1000+ 0.064 + 12(10.4)
= 1000 + 0.064 + 124.8 =1124.864
(iv) (598)3 = (600-2)3
= (600)3 – (2)3 – 3 × 600 × 2(600 – 2)
= 216000000 – 8-21,60,000 + 7200
=21,38,47,192
(v) (402)3 = (400 + 2)3
= (400)3 + (2)3 + 3 × 400 × 2(400 + 2)
= 64000000 + 8 + 960000 + 4800
= 64964808
(vi) (1002)3 = (1000 + 2)3
= (1000)3 + (2)3 + 3 × 1000 × 2(1000 + 2)
= 1000000000 + 8+ 6000000 + 12000
= 1006012008
प्रश्न 12.
निम्न के मान ज्ञात कीजिए।
(i) (46)3 (34)3
(ii) (23)3 – (17)3
(iii) (111)3 – (89)3
हल:
(i) (46)3 + (34)3
= (40 + 6)3 + (40-6)3
= (40)3 + (6)3 + 3 × 40 × 6(40 + 6) + (40)3 – (6)3 – 3 × 40 × 6(40 – 6)
= 2(40)3 + 3 × 40 × 6[46 – 34]
= 2(40)3 + 720 × 12
= 128000 + 8640 = 136640
(ii) (23)3 – (17)3
= (20 + 3)3 – (20 – 3)3
= (20)3 + (3)3 + 3 × 20 × 3(20 + 3) – (20)3 + (3)3 + 3 × 20 × 3(20 – 3)
= 2(3)3 + 3 × 20 × 3(23 + 17)
= 2 × 27 + 180(40) = 54+ 7200 = 7254
(iii) (111)3 – (89)3
= (100 + 11)3 – (100 – 11)3
= (100)3 + (11)3 + 3 × 100 × 11 (100 + 11) – (100)3 + (11)3 + 3 × 100 × 11(100 – 11)
= 2(11)3 + 3 × 100 × 11(111 + 89)
= 2(1331) + 3 × 100 × 11 × 200
= 2662 + 660000 = 662662
प्रश्न 13.
यदि 3x +2y = 20 व xy = [latex]\frac{14}{9}[/latex] तब 27x3 + 8y3 का मान ज्ञात कीजिए।
हलः
3x + 2y = 20 …………… (1)
वर्ग करने पर
9x2 + 4y2 + 12xy = 400
9x2 +4y2 = 400 – 12xy
27x3 + 8y3
= (3x)3 + (2y)3
= (3x + 2y) (9x2 + 4y2 – 6xy)
= (3x + 2y)(400 – 12xy – 6xy)
= (3x + 2y)(400 – 18xy) = (20)( 400 – 18 × [latex]\frac{14}{9}[/latex])
= (20) (400 – 28) = (20) (372) = 7440
प्रश्न 14.
हलः
प्रश्न 15.
हलः
प्रश्न 16.
हलः
