Question

Solve (ab+1)^2-(ab)^2 where, b-a=1

Answer 1

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$b - a = 1 \\ b = 1 + a$

${ab + 1}^{2} - {ab}^{2} \\ = (ab + 1 + ab)(ab + 1 - ab) \\ = (2ab + 1)(1) \\ \\ 2ab + 1 \\ = 2a(1 + a) + 1 \\ = 2a + 2 {a}^{2} + 1 \\ 2 {a}^{2} + 2a + 1 \\ = 2 {a}^{2} + 2a + 1a + 1 \\ = 2a(a + 1) + 1(a + 1) \\ (2a + 1)(a + 1) \\ \\ 2a + 1 = 0 \\ 2a = - 1 \\ a = \frac{ - 1}{2} \\ \\ a + 1 = 0 \\ a = - 1 \\ \\ fill \: value \: in \: first \: equation$

$b = 1 + \frac{ - 1}{2} \\ b = \frac{2 - 1}{2} \\ b = \frac{1}{2}$

$b = 1 + ( - 1) \\ b = 1 - 1 \\ b = 0$

hope it will help you

Answer 2

Answer:

here is your answer

Step-by-step explanation:

i think it may help you out