Question

1/1+root2 + 1/root2+root3 + 1/root3+root4

Answer 1

Answer:

6.72361463913

Step-by-step explanation:

Answer 2

$\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+2}$

First we may rationalise the denominator.

Consider  $\dfrac{1}{1+\sqrt{2}}.$

We multiply both the numerator and the denominator by  $\sqrt2-1.$

$\dfrac{1(\sqrt2-1)}{(1+\sqrt2)(\sqrt2-1)}=\dfrac{\sqrt2-1}{(\sqrt2+1)(\sqrt2-1)}=\dfrac{\sqrt2-1}{2-1}=\sqrt2-1$

Consider  $\dfrac{1}{\sqrt2+\sqrt3}.$

We multiply both the numerator and the denominator by  $\sqrt3-\sqrt2.$

$\dfrac{1(\sqrt3-1\sqrt2)}{(\sqrt2+\sqrt3)(\sqrt3-\sqrt2)}=\dfrac{\sqrt3-\sqrt2}{(\sqrt3+\sqrt2)(\sqrt3-\sqrt2)}=\dfrac{\sqrt3-\sqrt2}{3-2}=\sqrt3-\sqrt2$

Consider  $\dfrac{1}{\sqrt3+2}.$

We multiply both the numerator and the denominator by  $2-\sqrt3.$

$\dfrac{1(2-\sqrt3)}{(\sqrt3+2)(2-\sqrt3)}=\dfrac{2-\sqrt3}{(2+\sqrt3)(2-\sqrt3)}=\dfrac{2-\sqrt3}{4-3}=2-\sqrt3$

So,

$\begin{aligned}&\frac{1}{1+\sqrt2}+\frac{1}{\sqrt2+\sqrt3}+\frac{1}{\sqrt3+2}\\ \\ \Longrightarrow\ \ &(\sqrt2-1)+(\sqrt3-\sqrt2)+(2-\sqrt3)\\ \\ \Longrightarrow\ \ &\sqrt2-1+\sqrt3-\sqrt2+2-\sqrt3\\ \\ \Longrightarrow\ \ &2-1\\ \\ \Longrightarrow\ \ &\Large\bold{1}\end{aligned}$

Hence,  1  is the answer.