Advertisements
Advertisements
प्रश्न
Prove that:
`(64/125)^(-2/3)+1/(256/625)^(1/4)+(sqrt25/root3 64)=65/16`
Advertisements
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`
APPEARS IN
संबंधित प्रश्न
Simplify the following
`(2x^-2y^3)^3`
Simplify the following
`(4ab^2(-5ab^3))/(10a^2b^2)`
Solve the following equation for x:
`2^(x+1)=4^(x-3)`
If 49392 = a4b2c3, find the values of a, b and c, where a, b and c are different positive primes.
Find the value of x in the following:
`2^(5x)div2x=root5(2^20)`
Write the value of \[\sqrt[3]{125 \times 27}\].
The value of x − yx-y when x = 2 and y = −2 is
If a, m, n are positive ingegers, then \[\left\{ \sqrt[m]{\sqrt[n]{a}} \right\}^{mn}\] is equal to
Find:-
`125^((-1)/3)`
Simplify:
`7^(1/2) . 8^(1/2)`
