The correct option is B a + b a^2 b^2a b (we know that a^2 b^2=(a+b)(a b)) =(a b)(a+b)(a b) =(a+b) ...

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Standard IX

Mathematics

Laws of Exponents for Real Numbers

a2-b2a-b is e...

Question

$\frac{a^{2} - b^{2}}{a - b} i s e q u a l t o$

a + b

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a - b

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a^{2} + b^{2}

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(a - b)^{2}

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Solution

The correct option is B $a + b$
$\frac{a^{2} - b^{2}}{a - b}$ (we know that $a^{2} - b^{2} = (a + b) (a - b))$
$= \frac{(a - b) (a + b)}{(a - b)}$

$= (a + b)$

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Similar questions

\frac{\sqrt{a^{2} - b^{2}} + a}{\sqrt{a^{2} + b^{2}} + b} \div \frac{\sqrt{a^{2} + b^{2}} - b}{a - \sqrt{a^{2} - b^{2}}}

Q. If θ =

\frac{a}{b}

, then

\frac{(a sinθ - b cosθ)}{(a \sin θ + b cosθ)}

= ?
(a)

\frac{(a^{2} + b^{2})}{(a^{2} - b^{2})}

(b)

\frac{(a^{2} - b^{2})}{(a^{2} + b^{2})}

(c)

\frac{a^{2}}{(a^{2} + b^{2})}

(d)

\frac{b^{2}}{(a^{2} + b^{2})}

Which of the following is correct?
a) $(a - b)^{2} = a^{2} + 2 a b - b^{2}$
b) $(a - b)^{2} = a^{2} - 2 a b + b^{2}$
c) $(a - b)^{2} = a^{2} - b^{2}$
d) $(a + b)^{2} = a^{2} + 2 a b - b^{2}$

Q. Simplify:

\frac{a^{2} - {(b - c)}^{2}}{{(a + c)}^{2} - b^{2}} + \frac{b^{2} - {(a - c)}^{2}}{{(a + b)}^{2} - c^{2}} + \frac{c^{2} - {(a - b)}^{2}}{{(b + c)}^{2} - a^{2}}

Q. If AB = A and BA = B, where A and B are square matrices, then
(a) B² = B and A² = A
(b) B² ≠ B and A² = A
(c) A² ≠ A, B² = B
(d) A² ≠ A, B² ≠ B

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a2−b2a−b is equal to

The correct option is B a+ba2−b2a−b (we know that a2−b2=(a+b)(a−b))=(a−b)(a+b)(a−b)=(a+b)

$\frac{a^{2} - b^{2}}{a - b} i s e q u a l t o$

The correct option is B $a + b$
$\frac{a^{2} - b^{2}}{a - b}$ (we know that $a^{2} - b^{2} = (a + b) (a - b))$
$= \frac{(a - b) (a + b)}{(a - b)}$

$= (a + b)$