In this blog we are going to solve this Question

 \frac{\sqrt{10}+\sqrt{15}+\sqrt{20}}{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}

We can rewrite the above equation as

 \frac{\sqrt{2.5}+\sqrt{3.5}+\sqrt{5.4}}{\sqrt{2}+\sqrt{3}+\sqrt{3.2}+\sqrt{4.2}+\sqrt{4.4}}

= \frac{\sqrt{2}.\sqrt{5}+\sqrt{3}.\sqrt{5}+\sqrt{5}.\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{3}.\sqrt{2}+\sqrt{4}.\sqrt{2}+\sqrt{4}.\sqrt{4}}

We will take \sqrt{5} common from Numerator

= \frac{\sqrt{5}(\sqrt{2}+\sqrt{3}+\sqrt{4})}{\sqrt{2}+\sqrt{3}+\sqrt{3}.\sqrt{2}+\sqrt{4}.\sqrt{2}+\sqrt{4}.\sqrt{4}}

= \frac{\sqrt{5}(\sqrt{2}+\sqrt{3}+2)}{\sqrt{2}+\sqrt{3}+\sqrt{3}.\sqrt{2}+2.\sqrt{2}+4}

Now in Denominator we write 4 as 2 +2 .It is done to make factor easily
 \frac{\sqrt{5}(\sqrt{2}+\sqrt{3}+2)}{(\sqrt{2}+\sqrt{3}+2)+\sqrt{2}(\sqrt{3}+\sqrt{2}+2)}

Now we will take (\sqrt{2}+\sqrt{3}+2) common in denomintor

 \frac{\sqrt{5}(\sqrt{2}+\sqrt{3}+2)}{(\sqrt{2}+\sqrt{3}+2)(1+\sqrt{2})}

Now cancel out (\sqrt{2}+\sqrt{3}+2) in both numerator and denominator we will get

 \frac{\sqrt{5}}{(1+\sqrt{2})}

Now we will multiply both numerator and denominator with 1-\sqrt{2} we will get
 \frac{\sqrt{5}}{(1+\sqrt{2})} \left( \frac{1-\sqrt{2}}{1-\sqrt{2}}  \right)

On further Solving we will get
\frac{\sqrt{5}-\sqrt{5.2}}{1-\sqrt{2}+\sqrt{2}-2} = -(\sqrt{5}-\sqrt{10})

after multiplying negative sign outside we will get

\sqrt{10}-\sqrt{5}

And this our Answer \sqrt{10}-\sqrt{5} Hope you guys like it