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