Solve -ab-1-2 - -ab-2 where- b - a - 1-a-1-2 - -b-1-2b-2a-2 - b-2a-2 - -b-1-2

The correct option is C 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 + 2a ...

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Byju's Answer

Standard VIII

Mathematics

Meaning and Representation of Square Root

Solve (ab+1)^...

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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Similar questions

Q. Solve the following quadratic equations by factorization:

\frac{1}{2 a + b + 2 x} = \frac{1}{2 a} + \frac{1}{b} + \frac{1}{2 x}

Q. Solve each of the following quadratic equations:

\frac{1}{2 a + b + 2 x} = \frac{1}{2 a} + \frac{1}{b} + \frac{1}{2 x}

[CBSE 2013]

Q. In a trapezium ABCD, if E and F be the mid-point of the diagonals AC and BD respectively. Then, EF = ?
(a)

\frac{1}{2} A B

(b)

\frac{1}{2} C D

(c)

\frac{1}{2} (A B + C D)

(d)

\frac{1}{2} (A B - C D)

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

Q. Solve

(a b + 1)^{2}

(a b)^{2}

where, 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.