Math Olympiad Question

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

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