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Prove That: `(64/125)^(-2/3)+1/(256/625)^(1/4)+(Sqrt25/Root3 64)=65/16` - Mathematics

Prove that:

`(64/125)^(-2/3)+1/(256/625)^(1/4)+(sqrt25/root3 64)=65/16`

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Solution

We have to prove that `(64/125)^(-2/3)+1/(256/625)^(1/4)+(sqrt25/root3 64)=65/16`

Let x = `(64/125)^(-2/3)+1/(256/625)^(1/4)+(sqrt25/root3 64)`

`=2^(6xx(-2)/3)/5^(3xx(-2)/3)+1/(2^(8xx1/4)/5^(4xx1/4))+sqrt(5xx5)/root3 (4xx4xx4)`

`=2^-4/5^-2+1/(2^2/5)+5/4`

`=(1/2^4)/(1/5^2)+5/2^2+5/4`

`rArrx=1/16xx25/1+5/4+5/4=65/16`

By taking least common factor we get

`x=(25+20+20)/16=65/16`

Hence, `(64/125)^(-2/3)+1/(256/625)^(1/4)+(sqrt25/root3 64)=65/16`

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APPEARS IN

RD Sharma Mathematics for Class 9
Chapter 2 Exponents of Real Numbers
Exercise 2.2 | Q 3.7 | Page 24
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