The identity used in simplifying 101 x 99 is 

A) (a + b) (a – b) = a2 – b2 

B) (a – b)2 = a2 – 2ab + b2 

C) (a + b)2 = a2 + 2ab + b2 

D) (x + a) (x + b) = x2 + (a + b)x + ab

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2 Answers

Correct option is (A) (a + b) (a – b) = a2 – b2

\(101\times99\) = (100+1) (100-1)

Let a = 100, b = 1

Then, \(101\times99\) \(=(a+b)(a-b)=a^2-b^2=100^2-1^2\) = 10000 - 1 = 9999

Thus, the identity used in simplifying \(101\times99\) is \((a+b)(a-b)=a^2-b^2.\)

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A) (a + b) (a – b) = a2 – b2 

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