Question

Simplify this equation

Answer 1

Heya user,
Here is your answer !!

[ Refer to the picture attached for the answer ]

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Answer 2

$\underline{\underline{\bold{Question:}}}$

$\bold{\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt2+\sqrt3}+\dfrac{1}{\sqrt3+\sqrt4}+......+\dfrac{1}{\sqrt8+\sqrt9}}$

$\underline{\bold{Solution:}}$

$\bold{\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt2+\sqrt3}+\dfrac{1}{\sqrt3+\sqrt4}+......+\dfrac{1}{\sqrt8+\sqrt9}}$

$\underline{\bold{By\:Rationalizing\:it\:we\:get,}}$

$\bold{\dfrac{1{\times}(\sqrt2-1)}{(\sqrt{2}+1)(\sqrt2-1)}+\dfrac{1{\times}(\sqrt3-\sqrt2)}{(\sqrt3+\sqrt2)(\sqrt3-\sqrt2)}+\dfrac{1{\times}(\sqrt4-\sqrt3)}{(\sqrt4+\sqrt3)(\sqrt4-\sqrt3)}+......+\frac{1{\times}(\sqrt9-\sqrt8)}{(\sqrt9+\sqrt8)(\sqrt9-\sqrt8)}}$

$\boxed{\bold{a^2-b^2=(a+b)(a-b)}}$

$\bold{=\sqrt2-1+\sqrt3-\sqrt2+\sqrt4-\sqrt3+......+\sqrt9-\sqrt8}\\\\\\\bold{=\sqrt9-1.}$

$\bold{=3-1=2.}\\\\\\\boxed{\boxed{\bold{Answer=2.}}}$

$\underline{\bold{Identity:}}\\\\\\\bold{1.(a+b)^2=a^2+2ab+b^2}\\\\\\\bold{2.(a-b)^2=a^2-2ab-b^2}\\\\\\\bold{3.\:a^2+b^2=(a+b)^2-2ab}\\\\\\\bold{4.\:a^2+b^2=(a-b)^2+2ab}$