Question

prove that 1/3 - root 8- 1/ root 8-root 7 + 1/ root 7 - root 6 -1root 6 - root 5 + 1/ root 5 -2 =5​

Answer 1

Step-by-step explanation:

$lhs= \frac{1}{3 - \sqrt{8} } - \frac{1}{ \sqrt{8} - \sqrt{7} } + \frac{1}{ \sqrt{7} - \sqrt{6} } - \frac{1}{ \sqrt{6} - \sqrt{5} } + \frac{1}{ \sqrt{5} - 2} = </p><p>3 + \sqrt{8} - \sqrt{8} - \sqrt{7} + \sqrt{7 } + \sqrt{6} - \sqrt{6} - \sqrt{5} + \sqrt{5} + 2 </p><p>= 3 + 2 = 5 \\</p><p> rhs = 5 \\ \\ lhs = rhs \\hence \: verified$

Answer 2

Answer:

$Step-by-step explanation:</p><p></p><p>\begin{gathered}lhs= \frac{1}{3 - \sqrt{8} } - \frac{1}{ \sqrt{8} - \sqrt{7} } + \frac{1}{ \sqrt{7} - \sqrt{6} } - \frac{1}{ \sqrt{6} - \sqrt{5} } + \frac{1}{ \sqrt{5} - 2} = < /p > < p > 3 + \sqrt{8} - \sqrt{8} - \sqrt{7} + \sqrt{7 } + \sqrt{6} - \sqrt{6} - \sqrt{5} + \sqrt{5} + 2 < /p > < p > = 3 + 2 = 5 \\ < /p > < p > rhs = 5 \\ \\ lhs = rhs \\hence \: verified\end{gathered}lhs=3−81−8−71+7−61−6−51+5−21=</p><p>3+8−8−7+7+6−6−5+5+2</p><p>=3+2=5</p><p>rhs=5lhs=rhshenceverified</p><p></p><p>$

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