Advertisements
Advertisements
Question
Assuming that x, y, z are positive real numbers, simplify the following:
`(sqrtx)^((-2)/3)sqrt(y^4)divsqrt(xy^((-1)/2))`
Advertisements
Solution
We have to simplify the following, assuming that x, y, z are positive real numbers
Given `(sqrtx)^((-2)/3)sqrt(y^4)divsqrt(xy^((-1)/2))`
`=(x^(1/2))^(-2/3)(y^4)^(1/2)div(x xxy^(-1/2))^(1/2)`
`=(x^(1/2xx-2/3)xxy^(4xx1/2))/(x^(1/2)xxy^(-1/2xx1/2))`
`=(x^(-1/3)xxy^2)/(x^(1/2)xxy^(-1/4))`
By using the law of rational exponents, `a^mdiva^n=a^(m-n)` we have
`=x^(-1/3-1/2)xxy^(2+1/4)`
`=x^(-5/6)xxy^(9/4)`
`=1/x^(5/6)xxy^(9/4)`
`=y^(9/4)/x^(5/6)`
APPEARS IN
RELATED QUESTIONS
Simplify the following
`((x^2y^2)/(a^2b^3))^n`
If abc = 1, show that `1/(1+a+b^-1)+1/(1+b+c^-1)+1/(1+c+a^-1)=1`
Solve the following equation for x:
`2^(x+1)=4^(x-3)`
Prove that:
`(1/4)^-2-3xx8^(2/3)xx4^0+(9/16)^(-1/2)=16/3`
Show that:
`[{x^(a(a-b))/x^(a(a+b))}div{x^(b(b-a))/x^(b(b+a))}]^(a+b)=1`
If a and b are different positive primes such that
`((a^-1b^2)/(a^2b^-4))^7div((a^3b^-5)/(a^-2b^3))=a^xb^y,` find x and y.
Which of the following is (are) not equal to \[\left\{ \left( \frac{5}{6} \right)^{1/5} \right\}^{- 1/6}\] ?
If \[4x - 4 x^{- 1} = 24,\] then (2x)x equals
Find:-
`125^((-1)/3)`
Simplify:
`(3/5)^4 (8/5)^-12 (32/5)^6`
