The lowest form of a-b-a2-b2 is 1-a-b-

The correct option is A True Given that, (a b)/(a^2 b^2) where, (a^2 b^2) = (a + b)(a b) Cancelling (a b) from both numerator and denominator, The shor ...

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Byju's Answer

Standard VIII

Mathematics

Numbers in General Form

The lowest fo...

Question

The lowest form of $\frac{(a - b)}{(a^{2} - b^{2})}$ is $\frac{1}{(a + b)}$ .

True

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False

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Solution

The correct option is A
True

Given that, $\frac{(a - b)}{(a^{2} - b^{2})}$

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

Cancelling (a-b) from both numerator and denominator,

The shortest form is $\frac{1}{(a + b)}$

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Similar questions

The lowest form of $\frac{(a + b)}{a b} - \frac{2}{(a + b)}$ if ab=1.

The lowest form of the $\frac{a^{2} - b^{2}}{a^{3} - b^{3}}$ is___

Q. Calculate the lowest form of the expression:

\frac{b^{2} - 1}{1 - b^{2}} \times \frac{b}{b^{3}}

\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})}

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The lowest form of (a − b)(a2 − b2) is 1(a + b).

The correct option is A True Given that, (a − b)(a2 − b2) where, (a2 − b2) = (a + b)(a − b) Cancelling (a-b) from both numerator and denominator, The shortest form is 1(a + b)

The lowest form of $\frac{(a - b)}{(a^{2} - b^{2})}$ is $\frac{1}{(a + b)}$ .

The correct option is A
True

Given that, $\frac{(a - b)}{(a^{2} - b^{2})}$

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

Cancelling (a-b) from both numerator and denominator,

The shortest form is $\frac{1}{(a + b)}$