Class 8 RD Sharma解决方案-第六章代数表达式和恒等式–练习6.6 |套装1

问题1.将以下二项式的平方写为三项式：

(i)(x + 2) ²

解决方案：

x² + 2 (x) (2) + 22

x² + 4x + 4

(ii)(8a + 3b) ²

解决方案：

(8a)² + 2 (8a) (3b) + (3b)

64a² + 48ab + 9b²

(iii)(2m +1) ²

解决方案：

(2m)² + 2 (2m) (1) + 1²

4m² + 4m + 1

(iv)(9a +1/6) ²

解决方案：

(9a)² + 2 (9a) (1/6) + (1/6)²

81a² + 3a + 1/36

(ⅴ)(X + X ^2/2)2

解决方案：

(x)² + 2 (x) (x²/2) + (x²/2)²

x² + x³ + 1/4x⁴

(vi)(x / 4 – y / 3) ²

解决方案：

(x/4)² – 2 (x/4) (y/3) + (y/3)²

1/16x² – xy/6 + 1/9y²

(vii)(3倍-1/3倍) ²

解决方案：

(3x)² – 2 (3x) (1/3x) + (1/3x)²

9x² – 2 + 1/9x²

( viii)(x / y – y / x) ²

解决方案：

(x/y)² – 2 (x/y) (y/x) + (y/x)²

x²/y² – 2 + y²/x²

(ix)(3a / 2 – 5b / 4) ²

解决方案：

(3a/2)² – 2 (3a/2) (5b/4) + (5b/4)²

9/4a² – 15/4ab + 25/16b²

(x)(a ² b – bc ² ) ²

解决方案：

(a²b)² – 2 (a²b) (bc²) + (bc²)²

a⁴b⁴– 2a²b²c² + b²c⁴

(xi)(2a / 3b + 2b / 3a) ²

解决方案：

(2a/3b)² + 2 (2a/3b) (2b/3a) + (2b/3a)²

4a²/9b² + 8/9 + 4b²/9a²

(xii)(x ² – ay) ²

解决方案：

(x²)² – 2 (x²) (ay) + (ay)²

x⁴ – 2x²ay + a²y²

问题2.查找以下二项式的乘积：

i)(2x + y)(2x + y)

解决方案：

2x (2x + y) + y (2x + y)

4x² + 2xy + 2xy + y²

4x² + 4xy + y²

(ii)(a + 2b)(a – 2b)

解决方案：

a (a – 2b) + 2b (a – 2b)

a² – 2ab + 2ab – 4b²

a² – 4b²

(iii)(a ² + bc)(a ² – bc)

解决方案：

a² (a² – bc) + bc (a² – bc)

a⁴ – a²bc + bca² – b²c²

a⁴ – b²c²

(iv)(4x / 5 – 3y / 4)(4x / 5 + 3y / 4)

解决方案：

4x/5 (4x/5 + 3y/4) – 3y/4 (4x/5 + 3y/4)

16/25x² + 12/20yx – 12/20xy – 9y²/16

16/25x² – 9/16y²

(v)(2x + 3 / y)(2x – 3 / y)

解决方案：

2x (2x – 3/y) + 3/y (2x – 3/y)

4x² – 6x/y + 6x/y – 9/y²

4x² – 9/y²

(vi)(2a ³ + b ³ )(2a ³ – b ³ )

解决方案：

2a³ (2a³ – b³) + b³ (2a³ – b³)

4a⁶ – 2a³b³ + 2a³b³ – b⁶

4a⁶ – b⁶

(vii)(x ⁴ + 2 / x ² )(x ⁴ – 2 / x ² )

解决方案：

= x⁴ (x⁴ – 2/x²) + 2/x² (x⁴ – 2/x²)

= x⁸ – 2x² + 2x² – 4/x⁴

= (x⁸ – 4/x⁴)

(viii)(x ³ + 1 / x ³ )(x ³ – 1 / x ³ )

解决方案：

= x³ (x³ – 1/x³) + 1/x³ (x³ – 1/x³)

= x⁶ – 1 + 1 – 1/x⁶

= x⁶ – 1/x⁶

问题3.使用平方二项式的公式，计算以下内容：

(i)(102) ²

解决方案：

We can rewrite 102 as 100 + 2

(102)2 = (100 + 2)2

By simplification ,

(100 + 2)2 = (100)2 + 2 (100) (2) + 22

= 10000 + 400 + 4 = 10404

(ii)(99) ²

解决方案：

We can rewrite 99 as 100 – 1

 (99)² = (100 – 1)²

On simplification,

(100 – 1)² = (100)² – 2 (100) (1) + 1²

= 10000 – 200 + 1 = = 9801

(iii)(1001) ²

解决方案：

We can rewrite 1001 as 1000 + 1

(1001)² = (1000 + 1)²

On simplification ,

(1000 + 1)² = (1000)² + 2 (1000) (1) + 1²

= 1000000 + 2000 + 1 = 1002001

(iv)(999) ²

解决方案：

We can rewrite 999 as 1000 – 1

(999)² = (1000 – 1)²

By simplification,

(1000 – 1)² = (1000)2 – 2 (1000) (1) + 12

= 1000000 – 2000 + 1 = 998001

(v)(703) ²

解决方案：

We can rewrite 700 as 700 + 3

(703)² = (700 + 3)²

By simplification,

(700 + 3)2 = (700)2 + 2 (700) (3) + 32

= 490000 + 4200 + 9 = 494209

问题4.使用以下公式简化以下内容：(a – b)(a + b)= a ² – b ² ：

(i)(82) ² –(18) ²

解决方案：

Here we will use the formula 

(82)² – (18)² = (82 – 18) (82 + 18)

= 64 × 100

= 6400

(ii)(467) ² –(33)2 ²

解决方案：

We will using the formula (a – b) (a + b) = a² – b²

(467)² – (33)² = (467 – 33) (467 + 33)

= (434) (500)

= 217000

(iii)(79) ² –(69) ²

解决方案：

We will using the formula (a – b) (a + b) = a² – b²

(79)² – (69)² = (79 + 69) (79 – 69)

= (148) (10)

= 1480

(iv)197×203

解决方案：

We can rewrite 203 as 200 + 3 and 197 as 200 – 3

We will using the formula (a – b) (a + b) = a² – b²

197 × 203 = (200 – 3) (200 + 3)

= (200)² – (3)²

= 40000 – 9

= 39991

(v)113×87

解决方案：

We can rewrite 113 as 100 + 13 and 87 as 100 – 13

We can using the formula (a – b) (a + b) = a² – b²

113 × 87 = (100 – 13) (100 + 13)

= (100)² – (13)²

= 10000 – 169

= 9831

(vi)95×105

解决方案：

We can rewrite 95 as 100 – 5 and 105 as 100 + 5

We will  using the formula (a – b) (a + b) = a² – b²

95 × 105 = (100 – 5) (100 + 5)

= (100)² – (5)²

= 10000 – 25

= 9975

(vii)1.8×2.2

解决方案：

We can rewrite 1.8 as 2 – 0.2 and 2.2 as 2 + 0.2

We will using the formula (a – b) (a + b) = a² – b²

1.8 × 2.2 = (2 – 0.2) ( 2 + 0.2)

= (2)² – (0.2)²

= 4 – 0.04

= 3.96

(viii)9.8×10.2

解决方案：

We can rewrite 9.8 as 10 – 0.2 and 10.2 as 10 + 0.2

We will using the formula (a – b) (a + b) = a² – b²

9.8 × 10.2 = (10 – 0.2) (10 + 0.2)

= (10)² – (0.2)²

= 100 – 0.04

= 99.96

问题5.使用标识简化以下内容：

(i)(((58) ² –(42) ² )/ 16

解决方案：

We will using the formula (a – b) (a + b) = a² – b²

((58)² – (42)²)/16 = ((58-42) (58+42)/16)

= ((16) (100)/16)

= 100

(ii)178×178 – 22×22

解决方案：

We will using the formula (a – b) (a + b) = a² – b²

178 × 178 – 22 × 22 = (178)² – (22)²

= (178-22) (178+22)

= 200 × 156

= 31200

(iii)(198×198 – 102×102)/ 96

解决方案：

We using the formula (a – b) (a + b) = a² – b²

(198 × 198 – 102 × 102)/96 = ((198)2 – (102)2)/96

= ((198-102) (198+102))/96

= (96 × 300)/96

= 300

(iv)1.73×1.73 – 0.27×0.27

解决方案：

We will using the formula (a – b) (a + b) = a² – b²

1.73 × 1.73 – 0.27 × 0.27 = (1.73)2 – (0.27)2

= (1.73-0.27) (1.73+0.27)

= 1.46 × 2

= 2.92

(v)(8.63×8.63 – 1.37×1.37)/0.726

解决方案：

We will using the formula (a – b) (a + b) = a² – b²

(8.63 × 8.63 – 1.37 × 1.37)/0.726 = ((8.63)2 – (1.37)2)/0.726

= ((8.63-1.37) (8.63+1.37))/0.726

= (7.26 × 10)/0.726

= 72.6/0.726

= 100

问题6.在以下情况下，找到x的值：

(i)4x =(52) ² –(48) ²

解决方案：

We will using the formula (a – b) (a + b) = a² – b²

4x = (52)² – (48)²

4x = (52 – 48) (52 + 48)

4x = 4 × 100

4x = 400

x = 100

(ii)14x =(47) ² –(33) ²

解决方案：

We will using the formula (a – b) (a + b) = a2 – b2

14x = (47)2 – (33)2

14x = (47 – 33) (47 + 33)

14x = 14 × 80

x = 80

(iii)5倍=(50) ² –(40) ²

解决方案：

We using the formula (a – b) (a + b) = a2 – b2

5x = (50)² – (40)²

5x = (50 – 40) (50 + 40)

5x = 10 × 90

5x = 900

x = 180

问题7.如果x + 1 / x = 20，则求出x ² + 1 / x ^2的值。

解决方案：

Given equation in the x + 1/x = 20

when squaring both sides, we get

(x + 1/x)² = (20)²

x² + 2 × x × 1/x + (1/x)² = 400

x² + 2 + 1/x² = 400

x² + 1/x² = 398

问题8.如果x – 1 / x = 3，则找到x ² + 1 / x ²和x ⁴ + 1 / x ^4的值。

解决方案：

Given in the question x – 1/x = 3

 when squaring both sides,

(x – 1/x)² = (3)²

x² – 2 × x × 1/x + (1/x)² = 9

x² – 2 + 1/x² = 9

x² + 1/x² = 9+2

x² + 1/x² = 11

Now again when we square on both sides ,

(x² + 1/x²)² = (11)²

x⁴ + 2 × x² × 1/x² + (1/x²)² = 121

x⁴ + 2 + 1/x⁴ = 121

x⁴ + 1/x⁴ = 121-2

x⁴ + 1/x⁴ = 119

x² + 1/x² = 11

x⁴ + 1/x⁴ = 119

问题9.如果x ² + 1 / x ² = 18，则找到x + 1 / x和x – 1 / x的值。

解决方案：

Given in the question x² + 1/x² = 18

When adding 2 on both sides, 

x² + 1/x² + 2 = 18 + 2

x² + 1/x² + 2 × x × 1/x = 20

(x + 1/x)² = 20

x + 1/x = √20

When subtracting 2 from both sides, 

x²+ 1/x² – 2 × x × 1/x = 18 – 2

(x – 1/x)² = 16

x – 1/x = √16

x – 1/x = 4

问题10。如果x + y = 4且xy = 2，则求出x ² + y ^2的值

解决方案：

We know that x + y = 4 and xy = 2

Upon squaring on both sides of the given expression, we get

(x + y)² = 42

x² + y² + 2xy = 16

x² + y² + 2 (2) = 16   (since x y=2)

x² + y² + 4  = 16

x² + y² = 16 – 4

x² + y² =12

第6章代数表达式和身份–练习6.6 |套装2