9类RD Sharma解决方案–第4章代数恒等式-练习4.1 |套装1

问题1.使用身份评估以下各项：

(i)(2x – 1 / x) ²
(ii)(2x + y)(2x – y)
(iii)(a ² b – b ² a) ²
(iv)(a – 0.1)(a + 0.1)
(v)(1.5.x ² – 0.3y ² )(1.5x ² + 0.3y ² )

解决方案：

i) (2x – 1/x)² 

We know that, (a -b)² = a² – 2ab + b²

So, (2x – 1/x)² = (2x)² – (2 × 2x × 1/x) + (1/x)²

= 4x² + 1/x² – 4

ii) (2x + y) (2x – y)

We know that, (a + b)(a – b) = a² – b²

So, (2x + y) (2x – y) = (2x)² – (y)²

= 4x² – y²

iii) (a²b – b²a)²

We know that, (a -b)² = a² – 2ab + b²

So, (a²b – b²a)² = (a²b)² – (2 × a²b × b²a) + (b²a)²

= a⁴b² + b⁴a² – 2a³b³

iv) (a – 0.1) (a + 0.1)

We know that, (a + b)(a – b) = a² – b²

So, (a – 0.1) (a + 0.1) = (a)² – (0.1)²

= a² – 0.01

v) (1.5.x² – 0.3y²) (1.5x² + 0.3y²)

We know that, (a + b)(a – b) = a² – b²

So, (1.5.x² – 0.3y²) (1.5.x² + 0.3y²) = (1.5.x²)² – (0.3y²)²

= 2.225x⁴ – 0.09y⁴

问题2.使用身份评估以下各项：

(i)(399) ²
(ii)(0.98) ²
(iii)991×1009
(iv)117×83

解决方案：

i) (399)²

We can write (399)² as (400 – 1)²

Also, (a -b)² = a² – 2ab + b²

= (400)² + (1)² – 2 × 400 ×1

= 160000 + 1 – 800 = 159201

Hence, (399)² = 159201

ii) (0.98)²

We can write (0.98)² as (1 – 0.02)²

Also, (a -b)² = a² – 2ab + b²

= (1)² + (0.02)² – 2 × 0.02 ×1

= 1 + 0.0004 – 0.04 = 0.9604

Hence, (0.98)² = 0.9604

iii) 991 × 1009

We can write 991 × 1009 as (1000 – 9)(1000 + 9)

Also, (a + b)(a – b) = a² – b²

= (1000)² – (9)²

= 1000000 – 81 = 999919

Hence, 991 × 1009 = 999919

iv) 117 × 83

We can write 117 × 83 as (100 + 17)(100 – 17)

Also, (a + b)(a – b) = a² – b²

= (100)² – (17)²

= 10000 – 289 = 9711

Hence, 117 × 83 = 9711

问题3.简化以下各项：

(i)175×175 +2×175×25 + 25×25
(ii)322×322 – 2×322×22 + 22×22
(iii)0.76×0.76 + 2×0.76×0.24 + 0.24×0.24
(iv)(7.83×7.83 – 1.17×1.17)/6.66

解决方案：

i) 175 × 175 +2 × 175 × 25 + 25 × 25

It can be written as (175)² + 2(175)(25) + (25)²

And we also know that, (a + b)² = a² + b² + 2ab

So, we can conclude (175)² + 2(175)(25) + (25)² as (175 + 25)²

= (200)² = 40000

Hence, 175 × 175 +2 × 175 × 25 + 25 × 25 = 40000

ii) 322 × 322 – 2 × 322 × 22 + 22 × 22

It can be written as (322)² – 2(322)(22) + (22)²

And we also know that, (a – b)² = a² + b² – 2ab

So, we can conclude (322)² – 2(322)(22) + (22)² as (322 – 22)²

= (300)² = 90000

Hence, 322 × 322 – 2 × 322 × 22 + 22 × 22 = 90000

iii) 0.76 × 0.76 + 2 × 0.76 × 0.24 + 0.24 × 0.24

It can be written as (0.76)² + 2(0.76)(0.24) + (0.24)²

And we also know that, (a + b)² = a² + b² + 2ab

So, we can conclude (0.76)² + 2(0.76)(0.24) + (0.24)² as (0.76 + 0.24)²

= (1.0)² = 1

Hence, 0.76 × 0.76 + 2 × 0.76 × 0.24 + 0.24 × 0.24 = 1

iv) (7.83 × 7.83 – 1.17 × 1.17)/6.66

It can be written as (7.83² – 1.17²)/6.66

And we also know that, (a + b)(a – b) = a² – b²

So, we can conclude (7.83² – 1.17²)/6.66 as [(7.83 + 1.17)(7.83 – 1.17)]/6.66

= (9 × 6.66)/6.66 = 9

Hence, (7.83 × 7.83 – 1.17 × 1.17)/6.66 = 9

问题4.如果x + 1 / x = 11，则求出x ² + 1 / x ^2的值

解决方案：

Given, x + 1/x = 11

So, (x + 1/x)² = (x)² + (1/x)² + 2 × (x) × (1/x)

We also know that, (a + b)² = a² + b² + 2ab

So,

(11)² = x² + 1/x² + 2

121 – 2 = x² + 1/x²

x² + 1/x² = 119

Hence, value of x² + 1/x² is 119

问题5.如果x – 1 / x = -1，则求出x ²的值+ 1 / x ²

解决方案：

Given, x – 1/x = -1

So, (x – 1/x)² = (x)² + (1/x)² – 2 × (x) × (1/x)

We also know that, (a – b)² = a² + b² – 2ab

So,

(-1)² = x² + 1/x² – 2

1 + 2 = x² + 1/x²

x² + 1/x² = 3

Hence, value of x² – 1/x² is 3

问题6.如果x + 1 / x =√5，则求出x ² + 1 / x ²和x ⁴ + 1 / x ^4的值

解决方案：

Given, x + 1/x = √5

So, (x + 1/x)² = (x)² + (1/x)² + 2 × (x) × (1/x)

We also know that, (a + b)² = a² + b² + 2ab

So,

(√5)² = x² + 1/x² + 2

5 – 2 = x² + 1/x²

x² + 1/x² = 3

Now, taking the square of x² + 1/x²

(x² + 1/x²)² = x⁴ + 1/x⁴ + 2 × (x)² × (1/x²)

(3)² = x⁴ + 1/x⁴ + 2

x⁴ + 1/x⁴ = 7

Hence, value of x² + 1/x² is 3 and that of x⁴ +1/x⁴ is 7

问题7.如果x ² + 1 / x ² = 66，则求出x – 1 / x的值

解决方案：

Given, x² +1/x² = 66

Let us take the square of x – 1/x

So, (x – 1/x)² = (x)² + (1/x)² – 2 × (x) × (1/x)

Since, (a – b)² = a² + b² – 2ab

So,

(x – 1/x)² = 66 – 2

(x – 1/x)² = 64

x – 1/x = ±8

Hence, the value of x – 1/x is 8