Question

prove that:in this image
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Answer 1

$\Large{\underline{\underline{\mathfrak{\bf{Question}}}}}$

To prove:-

$\tt{\red{\:\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+........+\dfrac{1}{\sqrt{8}+\sqrt{9}}\:=\:2}}$

$\Large{\underline{\underline{\mathfrak{\bf{Prove}}}}}$

Teke, L.H.S.,

$:\mapsto\tt{\orange{\:\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+........+\dfrac{1}{\sqrt{8}+\sqrt{9}}}} \\ \\ \\ \\ \small\tt{\green{\:\:\:Rationalize\:of\:denominator\:of\:all\:term}} \\ \\ \\ \\ :\mapsto\tt{\:\dfrac{(1-\sqrt{2})}{(1+\sqrt{2})(1-\sqrt{2})}+\dfrac{(\sqrt{2}-\sqrt{3})}{(\sqrt{2}+\sqrt{3})(\sqrt{2}-\sqrt{3})}+.........+\dfrac{(\sqrt{8}-\sqrt{9})}{(\sqrt{8}-\sqrt{9})(\sqrt{8}+\sqrt{9})}} \\ \\ \\ \\ :\mapsto\tt{\:\dfrac{(1-\sqrt{2})}{(1^2-\sqrt{2}^2)}+\dfrac{(\sqrt{2}-\sqrt{3})}{(\sqrt{2}^2-\sqrt{3}^2)}+..........+\dfrac{(\sqrt{8}-\sqrt{9})}{(\sqrt{8}^2-\sqrt{9}^2)}} \\ \\ \\ \\ :\mapsto\tt{\:\dfrac{(1-\sqrt{2})}{1-2}+\dfrac{(\sqrt{2}-\sqrt{3})}{2-3}+........+\dfrac{(\sqrt{8}-\sqrt{9})}{8-9}} \\ \\ \\ \\ :\mapsto\tt{\:\dfrac{(1-\sqrt{2})}{-1}+\dfrac{(\sqrt{2}-\sqrt{3})}{-1}+........+\dfrac{(\sqrt{8}-\sqrt{9})}{-1}} \\ \\ \\ \\ :\mapsto\tt{\:(\sqrt{2}-1)+(\sqrt{3}-\sqrt{2})+......+(\sqrt{9}-\sqrt{8})}$

Here, only two term -1 and √9 are still .

Other, terms subtract.

So,

$:\mapsto\tt{\:-1+\sqrt{9}} \\ \\ \\ :\mapsto\tt{\:-1+3} \\ \\ \\ :\mapsto\tt{\orange{\:2}}$

= R.H.S.

That's proved.

Answer 2

Step-by-step explanation:

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$\bf \huge \: Question \: \:$

• Prove that 1/1+√2+1/√2+√3 + ........+1/√8+√9=2

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$\bf \huge \: To\: Find\:$

• Prove that 1/1+√2+1/√2+√3 + ........+1/√8+√9=2

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$\bf \huge \: Solution\:\:$

1/1+√2+1/√2+√3 + ........+1/√8+√9=2

1/2+√2= (1/1+√2)(√2-1/√2-1)

Since by rationalisation

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√2 -1/1

We have ,

similarly for all

Solving,

Supposed:-

-1 +√9

We know ,

√9 =3

Then putting the value,

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-1+3

= 2

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Hence PROVED,