Math, asked by burhan331, 11 months ago

Evaluate each of the following using identities:
(i) (2x - 1/x)²
(ii) (2x+y) (2x-y)
(iii) (a²b - b²a)²
(iv) (a – 0.1) (a +0.1)
(v)(1.5x² - 0.3y²)(1.5x² - 0.3y²)

Answers

Answered by nikitasingh79
18

(i) (2x - 1/x)²

 By using identity: (a – b)² = a² + b² – 2ab :  

(2x - 1/x)² = (2x)² + (1/x)² – 2 (2x)(1/x)

= 4x² + 1/x² – 4

 

(ii) (2x + y) (2x – y)

By using identity : (a – b)(a + b) = a² – b²  

(2x + y) (2x – y) = (2x )² – (y)²

= 4x² – y²

 

(iii) (a²b - b²a)²

By using identity: (a – b)² = a² + b² – 2ab  

(a²b - b²a)² = (a²b) ² + (b²a)² – 2(a²b)( b²a)

= a⁴b² +  b⁴a²– 2 a³b³

 

(iv) (a – 0.1) (a + 0.1)

By using identity: (a – b)(a + b) = a²– b²

(a – 0.1) (a + 0.1) = (a)² – (0.1)²

= a² – 0.01

 

(v) (1.5 x² – 0.3y²) (1.5 x² + 0.3y²)

By using identity: (a – b)(a + b) = a²– b²

(1.5 x² – 0.3y²) (1.5x² + 0.3y²) = (1.5x² ) ²– (0.3y²)²

= 2.25 x⁴ – 0.09y⁴

HOPE THIS ANSWER WILL HELP YOU…..

 

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Answered by Anonymous
10

Answer:

Step-by-step explanation:

(i) (2x - 1/x)²

 By using identity: (a – b)² = a² + b² – 2ab :  

(2x - 1/x)² = (2x)² + (1/x)² – 2 (2x)(1/x)

= 4x² + 1/x² – 4

 

(ii) (2x + y) (2x – y)

By using identity : (a – b)(a + b) = a² – b²  

(2x + y) (2x – y) = (2x )² – (y)²

= 4x² – y²

 

(iii) (a²b - b²a)²

By using identity: (a – b)² = a² + b² – 2ab  

(a²b - b²a)² = (a²b) ² + (b²a)² – 2(a²b)( b²a)

= a⁴b² +  b⁴a²– 2 a³b³

 

(iv) (a – 0.1) (a + 0.1)

By using identity: (a – b)(a + b) = a²– b²

(a – 0.1) (a + 0.1) = (a)² – (0.1)²

= a² – 0.01

 

(v) (1.5 x² – 0.3y²) (1.5 x² + 0.3y²)

By using identity: (a – b)(a + b) = a²– b²

(1.5 x² – 0.3y²) (1.5x² + 0.3y²) = (1.5x² ) ²– (0.3y²)²

= 2.25 x⁴ – 0.09y⁴

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