Question

evaluate 15/(√10+√20+√40-√5-√80)? plc now

Answer 1

Answer:

After Evaluating we get √10 + √5

Step-by-step explanation:

Given Expression;

$\frac{15}{\sqrt{10}+\sqrt{20}+\sqrt{40}-\sqrt{5}-\sqrt{80}}$

We need to evaluate the given expression.

We first simplify the square roots.

Consider,

$\frac{15}{\sqrt{10}+\sqrt{20}+\sqrt{40}-\sqrt{5}-\sqrt{80}}$

$=\frac{15}{\sqrt{10}+\sqrt{4\times5}+\sqrt{4\times10}-\sqrt{5}-\sqrt{16\times5}}$

$=\frac{15}{\sqrt{10}+2\sqrt{5}+2\sqrt{10}-\sqrt{5}-4\sqrt{5}}$

$=\frac{15}{(1+2)\sqrt{10}+(2-1-4)\sqrt{5}}$

$=\frac{15}{3\sqrt{10}-3\sqrt{5}}$

$=\frac{15}{3(\sqrt{10}-\sqrt{5})}$

$=\frac{5}{\sqrt{10}-\sqrt{5}}$

$=\frac{5}{\sqrt{10}-\sqrt{5}}\times\frac{\sqrt{10}+\sqrt{5}}{\sqrt{10}+\sqrt{5}}$

$=\frac{5(\sqrt{10}+\sqrt{5})}{(\sqrt{10}-\sqrt{5})(\sqrt{10}+\sqrt{5})}$

$=\frac{5(\sqrt{10}+\sqrt{5})}{(\sqrt{10})^2-(\sqrt{5})^2}$

$=\frac{5(\sqrt{10}+\sqrt{5})}{10-5}$

$=\frac{5(\sqrt{10}+\sqrt{5})}{5}$

$=\sqrt{10}+\sqrt{5}$

Therefore, After Evaluating we get √10 + √5