Question

The value of $\left[\frac{a^2-5ab}{a^2-6ab+5a{b}^2}\times \frac{a^2-b^2}{a^2+ab}\right]$ is :

• A. $-1$
• B. $\frac{a}{b}$
• C. $\frac{1}{a}$
• D. $1$

Answer 1

The correct option is D

$1$

Solution:

Find the value of given expression:

Given: The expression $\left[\frac{a^2-5ab}{a^2-6ab+5a{b}^2}\times \frac{a^2-b^2}{a^2+ab}\right]$

$$\begin{gathered} \left[\frac{a^2-5ab}{a^2-6ab+5a{b}^2}\times \frac{a^2-b^2}{a^2+ab}\right] \\ =\frac{a(a-5b)\times \left(a+b\right)\left(a-b\right)}{\left(a^2-6ab+5b^2\right)\times a\left(a+b\right)}\mathbf{\left[}\mathbf{\because }{\mathbf{a}}^{\mathbf{2}}\mathbf{-}{\mathbf{b}}^{\mathbf{2}}\mathbf{=}\mathbf{(}\mathbf{a}\mathbf{+}\mathbf{b}\mathbf{)}\mathbf{\times }\mathbf{(}\mathbf{a}\mathbf{-}\mathbf{b}\mathbf{)}\mathbf{\right]}\mathbf{}[\mathrm{Take}\text{'}\mathrm{a}\text{'}\mathrm{common}\mathrm{from}({\mathrm{a}}^2-5\mathrm{ab})\mathrm{and}({\mathrm{a}}^2+\mathrm{ab})] \\ =\frac{\left(a-5b\right)\left(a-b\right)}{a^2-5ab-ab+5b^2}(\mathrm{factorise}\mathrm{the}\mathrm{denominator}) \\ =\frac{(a-5b)\left(a-b\right)}{a\left(a-5b\right)-b(a-5b)} \\ =\frac{\left(a-5b\right)(a-b)}{(a-5b)\left(a-b\right)}(\mathrm{taking}\mathrm{common}\mathrm{term}(\mathrm{a}-5\mathrm{b}\left)\right) \\ =\mathbf{1} \end{gathered}$$

Final answer: Hence, option (D) is correct.