9类NCERT解决方案–第2章多项式–练习2.5 |套装2

第2章多项式–练习2.5 |套装1

问题9.验证：

(i)x ³ + y ³ =(x + y)(x ² – xy + y ² )

解决方案：

Formula (x + y)³ = x³ + y³ + 3xy(x + y)

x³ + y³ = (x + y)³ – 3xy(x + y)

x³ + y³ = (x + y) [(x + y)² – 3xy]

x³ + y³ = (x + y) [(x² + y² + 2xy) – 3xy]

Therefore, x³ + y³ = (x + y) (x² + y² – xy)

(ii)x ³ – y ³ =(x – y)(x ² + xy + y ² )

解决方案：

Formula, (x – y)³ = x³ – y³ – 3xy(x – y)

x³ − y³ = (x – y)³ + 3xy(x – y)

x³− y³ = (x – y) [(x – y)² + 3xy]

 x³ − y³ = (x – y) [(x² + y² – 2xy) + 3xy]

Therefore, x³ + y³ = (x – y) (x² + y² + xy)

问题10：分解以下各项：

(i)27y ³ + 125z ³

解决方案：

27y³ + 125z³ can also be written as (3y)³ + (5z)³

27y³ + 125z³ = (3y)³ + (5z)³

Formula x³ + y³ = (x + y) (x² – xy + y²)

27y³ + 125z³ = (3y)³ + (5z)³

= (3y + 5z) [(3y)² – (3y)(5z) + (5z)²]

= (3y + 5z) (9y² – 15yz + 25z²)

(ii)64m ³ – 343n ³

解决方案：

64m³ – 343n³ can also be written as (4m)³ – (7n)³

64m³ – 343n³ = (4m)3 – (7n)³

Formula x³ – y³ = (x – y) (x² + xy + y²)

64m³ – 343n³ = (4m)³ – (7n)³

= (4m – 7n) [(4m)² + (4m)(7n) + (7n)²]

问题11：分解：27x ³ + y ³ + z ³ – 9xyz

解决方案：

27x³ + y³ + z³ – 9xyz can also be written as (3x)³ + y³ + z³ – 3(3x)(y)(z)

27x³ + y³ + z³ – 9xyz  = (3x)³ + y³ + z³ – 3(3x)(y)(z)

Formula, x³ + y³ + z³ – 3xyz = (x + y + z) (x² + y² + z² – xy  – yz – zx)

27x³ + y³ + z³ – 9xyz  = (3x)³ + y³ + z³ – 3(3x)(y)(z)

= (3x + y + z) [(3x)² + y² + z² – 3xy – yz – 3xz]

= (3x + y + z) (9x² + y² + z² – 3xy – yz – 3xz)

问题12.验证：x ³ + y ³ + z ³ – 3xyz =(1/2)(x + y + z)[(x – y) ² +(y – z) ² +(z – x) ² ]

解决方案：

Formula, x³ + y³ + z³ − 3xyz = (x + y + z)(x² + y² + z² – xy – yz – xz)

Multiplying by 2 and dividing by 2

= (1/2) (x + y + z) [2(x² + y² + z² – xy – yz – xz)]

= (1/2) (x + y + z) (2x² + 2y² + 2z² – 2xy – 2yz – 2xz)

= (1/2) (x + y + z) [(x² + y² − 2xy) + (y² + z² – 2yz) + (x² + z² – 2xz)]

= (1/2) (x + y + z) [(x – y)² + (y – z)² + (z – x)²]

Therefore, x³ + y³ + z³ – 3xyz  = (1/2) (x + y + z) [(x – y)² + (y – z)² + (z – x)²]

问题13。如果x + y + z = 0，则表明x ³ + y ³ + z ³ = 3xyz。

解决方案：

Formula, x³ + y³ + z³ – 3xyz = (x + y + z) (x² + y² + z² – xy – yz – xz)

Given, (x + y + z) = 0,

Then, x³ + y³ + z³ – 3xyz = (0) (x² + y² + z² – xy – yz – xz)

x³ + y³ + z³ – 3xyz = 0

Therefore, x³ + y³ + z³ = 3xyz

问题14：在不实际计算多维数据集的情况下，找到以下每个值：

(i)(−12) ³ +(7) ³ +(5) ³

解决方案：

Let,

x = −12

y = 7

z = 5

We know that if x + y + z = 0, then x³ + y³ + z³ = 3xyz.

and we have −12 + 7 + 5 = 0

Therefore, (−12)³ + (7)³ + (5)³ = 3xyz

= 3 × -12 × 7 × 5

= -1260

(ii)(28) ³ +(−15) ³ +(−13) ³

解决方案：

Let, 

x = 28

y = −15

z = −13

We know that if x + y + z = 0, then x³ + y³ + z³ = 3xyz.

and we have, x + y + z = 28 – 15 – 13 = 0

Therefore, (28)³ + (−15)³ + (−13)³ = 3xyz

= 3 (28) (−15) (−13)

= 16380

问题15.给出以下每个矩形的长度和宽度的可能表达式，并给出其面积：

(i)面积：25a ² – 35a + 12

解决方案：

Using splitting the middle term method,

25a² – 35a + 12

25a² – 35a + 12 = 25a² – 15a − 20a + 12

= 5a(5a – 3) – 4(5a – 3)

= (5a – 4) (5a – 3)

Possible expression for length & breadth is  = (5a – 4)  & (5a  – 3)

(ii)区域：35y ² + 13y – 12

解决方案：

Using the splitting the middle term method,

35y² + 13y – 12 = 35y² – 15y + 28y – 12

= 5y(7y – 3) + 4(7y – 3)

= (5y + 4) (7y – 3)

Possible expression for length  & breadth is = (5y + 4) & (7y – 3)

问题16.下面给出体积的长方体的尺寸的可能表达式是什么?

(i)音量：3x ² – 12x

解决方案：

3x² – 12x can also be written as 3x(x – 4) 

= (3) (x) (x – 4) 

Possible expression for length, breadth & height = 3, x & (x – 4)

(ii)交易量：12ky ² + 8ky – 20k

解决方案：

12ky² + 8ky – 20k can also be written as 4k (3y² + 2y – 5)

12ky² + 8ky– 20k = 4k(3y² + 2y – 5)

Using splitting the middle term method.

= 4k (3y2 + 5y – 3y – 5)

= 4k [y(3y + 5) – 1(3y + 5)]

= 4k (3y + 5) (y – 1)

Possible expression for length, breadth & height= 4k, (3y + 5) & (y – 1)