Advertisements
Advertisements
Question
Assuming that x, y, z are positive real numbers, simplify the following:
`sqrt(x^3y^-2)`
Solution
We have to simplify the following, assuming that x, y, z are positive real numbers
Given `sqrt(x^3y^-2)`
As x and y are positive real numbers then we can write
`sqrt(x^3y^-2)=(x^3y^-2)^(1/2)`
`=(x^(3xx1/2)xxy^(-2xx1/2))`
`=(x^(3/2)y^-1)`
By using law of rational exponents `a^-n=1/a^n` we have
`sqrt(x^3y^-2)=x^(3/2)xx1/y`
`=x^(3/2)/y`
Hence the simplified value of `sqrt(x^3y^-2)` is `x^(3/2)/y`
APPEARS IN
RELATED QUESTIONS
Simplify the following
`(4ab^2(-5ab^3))/(10a^2b^2)`
Prove that:
`(a^-1+b^-1)^-1=(ab)/(a+b)`
Show that:
`1/(1+x^(a-b))+1/(1+x^(b-a))=1`
If 2x = 3y = 6-z, show that `1/x+1/y+1/z=0`
Solve the following equation:
`4^(2x)=(root3 16)^(-6/y)=(sqrt8)^2`
Which one of the following is not equal to \[\left( \frac{100}{9} \right)^{- 3/2}\]?
If \[4x - 4 x^{- 1} = 24,\] then (2x)x equals
\[\frac{5^{n + 2} - 6 \times 5^{n + 1}}{13 \times 5^n - 2 \times 5^{n + 1}}\] is equal to
The simplest rationalising factor of \[2\sqrt{5}-\]\[\sqrt{3}\] is
If \[\sqrt{2} = 1 . 414,\] then the value of \[\sqrt{6} - \sqrt{3}\] upto three places of decimal is
