Question

how do I do it please help me by solving this..​

Answer 1

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Answer 2

First the denominator has to be simplified much.

$\displaystyle \frac{25}{\sqrt{40}-\sqrt{80}} \\ \\ \\ \\ \frac{25}{\sqrt{4 \times 10}-\sqrt{16 \times 5}} \\ \\ \\ \\ \frac{25}{2\sqrt{10}-4\sqrt{5}} \\ \\ \\ \\ \frac{25}{2(\sqrt{10}-2\sqrt{5})}$

Now, rationalising the denominator...

$\displaystyle \frac{25(\sqrt{10}+2\sqrt{5})}{2(\sqrt{10}-2\sqrt{5})(\sqrt{10}+2\sqrt{5})} \\ \\ \\ \\ \frac{25(\sqrt{10}+2\sqrt{5})}{2((\sqrt{10})^2-(2\sqrt{5})^2)} \\ \\ \\ \\ \frac{25(\sqrt{10}+2\sqrt{5})}{2(100-20)} \\ \\ \\ \\ \frac{25(\sqrt{10}+2\sqrt{5})}{2 \times 80} \\ \\ \\ \\ \frac{5(\sqrt{10}+2\sqrt{5})}{2 \times 16}$

Now, putting the values of both √5 and √10,...

$\displaystyle \frac{5(3.162+2 \times 2.236)}{2 \times 16} \\ \\ \\ \\ \frac{5(3.162+4.472)}{32} \\ \\ \\ \\ \frac{5 \times 7.634}{32} \\ \\ \\ \\ \frac{5 \times 3.817}{16} \\ \\ \\ \\ \frac{19.085}{16} \\ \\ \\ \\ \Large \bold{1.1928125 \ \approx\ 1.193}$

Thus the given fraction is approximately evaluated!!!