Solve a b-12-a b2 where- b-a-1-A- a2-b-12B- a2-b2C- a-12-b-12D- b2

The correct option is B a^2 + b^2 (ab+1)^2 (ab)^2 = (a(a+1) + 1)^2 (a(a+1))^2 = (a^2(a+1)^2 +1 + 2a(a+1)) (a^2(a+1)^2) = 2a(a+1)+ 1 = 2a^2+2a+1 = a^2 + 2 ...

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Byju's Answer

Standard VIII

Mathematics

Integral and Positive Square Root

Solve a b+12-...

Question

Solve $(a b + 1)^{2}$ - $(a b)^{2}$ where, b - a = 1.

$a^{2}$ + $b^{2}$

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$a^{2}$ + $(b + 1)^{2}$

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$(a + 1)^{2}$ + $(b + 1)^{2}$

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$b^{2}$

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Solution

The correct option is A
$a^{2}$ + $b^{2}$

$(a b + 1)^{2} - (a b)^{2} = (a (a + 1) + 1)^{2} - (a (a + 1))^{2}$
$= (a^{2} (a + 1)^{2} + 1 + 2 a (a + 1)) - (a^{2} (a + 1)^{2})$
$= 2 a (a + 1) + 1$
$= 2 a^{2} + 2 a + 1$
$= a^{2} + 2 a + 1 + a^{2}$
$= (a + 1)^{2} + a^{2}$
$= b^{2} + a^{2}$ [since $b = a + 1$ ]

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

The correct option is A a2 + b2 (ab+1)2−(ab)2=(a(a+1)+1)2−(a(a+1))2 =(a2(a+1)2+1+2a(a+1))−(a2(a+1)2) =2a(a+1)+1 =2a2+2a+1 =a2+2a+1+a2 =(a+1)2+a2 =b2+a2 [since b=a+1]

Solve $(a b + 1)^{2}$ - $(a b)^{2}$ where, b - a = 1.