This is a class 10 Question From Simplification of Rational Expressions chapter of Unit Algebra (Mathematics).
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Solution:
Given,
= $\dfrac{2}{a-1} -\dfrac{a}{a²+1} -\dfrac{1}{a+1} -\dfrac{3}{a²-1}$
= $\dfrac{2}{a-1} -\dfrac{1}{a+1} -\left ( \dfrac{a}{a²+1} +\dfrac{3}{a²-1} \right )$
= $\dfrac{2(a+1) -1(a-1)}{(a-1)(a+1)} -\left ( \dfrac{a(a²-1) +3(a²+1)}{(a²+1)(a²-1)} \right )$
= $\dfrac{2a +2-a +1}{a²-1} - \left ( \dfrac{a³-a+3a²+3}{a⁴-1} \right )$
= $\dfrac{a+3}{a²-1} - \dfrac{a³+3a²-a+3}{a⁴-1}$
= $\dfrac{(a+3)(a²+1) - a³-3a²+a-3}{a⁴-1}$
= $\dfrac{a³+a+3a²+3 -a³-3a²+a-3}{a⁴-1}$
= $\dfrac{2a}{a⁴-1}$
Given,
= $\dfrac{2}{a-1} -\dfrac{a}{a²+1} -\dfrac{1}{a+1} -\dfrac{3}{a²-1}$
= $\dfrac{2}{a-1} -\dfrac{1}{a+1} -\left ( \dfrac{a}{a²+1} +\dfrac{3}{a²-1} \right )$
= $\dfrac{2(a+1) -1(a-1)}{(a-1)(a+1)} -\left ( \dfrac{a(a²-1) +3(a²+1)}{(a²+1)(a²-1)} \right )$
= $\dfrac{2a +2-a +1}{a²-1} - \left ( \dfrac{a³-a+3a²+3}{a⁴-1} \right )$
= $\dfrac{a+3}{a²-1} - \dfrac{a³+3a²-a+3}{a⁴-1}$
= $\dfrac{(a+3)(a²+1) - a³-3a²+a-3}{a⁴-1}$
= $\dfrac{a³+a+3a²+3 -a³-3a²+a-3}{a⁴-1}$
= $\dfrac{2a}{a⁴-1}$
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Question: Simplify: 2/(a-1) -a/(a²+1) -1/(a+1) -3/(a²-1) | SciPiPupil
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