Question

1 /Root4 + root 5 + 1 / root 5 + root 6 + 1 divided by root 5 + root 6 + 1 / root 7 plus root 8 + 1 / root 8 + root 9 equal to 1prove that 1 / root for + root 5 + 1 / root 5 + root 6 + 1 / root 6 + root 7 + 1 / root 7 + 28 + 1 / root 8 + 29 equal to 1

Answer 1

Answer:

We have to prove:

$\dfrac{1}{\sqrt{4}+\sqrt{5}}+\dfrac{1}{\sqrt{5} \sqrt{6}}+\dfrac{1}{\sqrt{6}+ \sqrt{7}}+\dfrac{1}{\sqrt{7} \sqrt{8}}+\dfrac{1}{\sqrt{8} +\sqrt{9}}$

on multiplying and dividing by conjugate terms we get:

$=\dfrac{\sqrt{4}-\sqrt{5}}{4-5}+\dfrac{\sqrt{5} -\sqrt{6}}{5-6}+\dfrac{\sqrt{6}- \sqrt{7}}{6-7}+\dfrac{\sqrt{7}- \sqrt{8}}{7-8}+\dfrac{\sqrt{8}- \sqrt{9}}{8-9}\\\\=\dfrac{\sqrt{4}-\sqrt{5}}{-1}+\dfrac{\sqrt{5} -\sqrt{6}}{-1}+\dfrac{\sqrt{6}- \sqrt{7}}{-1}+\dfrac{\sqrt{7}- \sqrt{8}}{-1}+\dfrac{\sqrt{8}- \sqrt{9}}{-1}\\\\\\=\dfrac{\sqrt{4}-\sqrt{5}+\sqrt{5}-\sqrt{6}+\sqrt{6}-\sqrt{7}+\sqrt{7}-\sqrt{8}+\sqrt{8}-\sqrt{9}}{-1}$

(since on cancelling the like terms but with opposite signs we get)

$=\dfrac{\sqrt{4}-\sqrt{9}}{-1}$

Now as we know:

$\sqrt{4}=2, \sqrt{9}=3$

we have:

$=\dfrac{2-3}{-1}\\\\=\dfrac{-1}{-1}\\\\=1$

Hence,

$\dfrac{1}{\sqrt{4}+\sqrt{5}}+\dfrac{1}{\sqrt{5} \sqrt{6}}+\dfrac{1}{\sqrt{6}+ \sqrt{7}}+\dfrac{1}{\sqrt{7} \sqrt{8}}+\dfrac{1}{\sqrt{8} +\sqrt{9}}=1$

Answer 2

Answer:Hope it will help you..

Step-by-step explanation:

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