Question

prove that 1/2+ root 3 +2/root5-root 3+1/2-root 5

Answer 1

☜☆☞ hey friend ☜☆☞


here is your answer ☆☞

hope it will help you

Devil_king ▄︻̷̿┻̿═━一

Answer 2

Answer:

The equation $\frac{1}{2+\sqrt{3}}+\frac{2}{\sqrt{5}-\sqrt{3}}+\frac{1}{2-\sqrt{5}}$  is 0.

Step-by-step explanation:

Given:

$\frac{1}{2+\sqrt{3}}+\frac{2}{\sqrt{5}-\sqrt{3}}+\frac{1}{2-\sqrt{5}}$

To find:

the above equation

Step 1

Let,

$\frac{1}{2+\sqrt{3}}+\frac{2}{\sqrt{5}-\sqrt{3}}+\frac{1}{2-\sqrt{5}}$

$=& \frac{1}{2+\sqrt{3}} \times \frac{2-\sqrt{3}}{2-\sqrt{3}}+\frac{2}{\sqrt{5}-\sqrt{3}} \times \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}+\sqrt{3}} \\ &+\frac{1}{2-\sqrt{5}} \times \frac{2+\sqrt{5}}{2+\sqrt{5}}$

Simplifying the above equation, we get

$=& \frac{2-\sqrt{3}}{4-3}+\frac{2(\sqrt{5}+\sqrt{3})}{5-3}+\frac{2+\sqrt{5}}{4-5}$

Step 2

By using the PEMDAS rule, we get

$={2-\sqrt{3}+\frac{2 \sqrt{5}+2 \sqrt{3}}{5-3}-(2+\sqrt{5})}{2}$

$=\frac{4-2 \sqrt{3}+2 \sqrt{5}+2 \sqrt{3}-4-2 \sqrt{5}}{2}$

= 0.

Therefore, the equation $\frac{1}{2+\sqrt{3}}+\frac{2}{\sqrt{5}-\sqrt{3}}+\frac{1}{2-\sqrt{5}}$  is 0.

#SPJ2