Advertisements
Advertisements
प्रश्न
Assuming that x, y, z are positive real numbers, simplify the following:
`sqrt(x^3y^-2)`
Advertisements
उत्तर
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
संबंधित प्रश्न
Prove that:
`1/(1 + x^(b - a) + x^(c - a)) + 1/(1 + x^(a - b) + x^(c - b)) + 1/(1 + x^(b - c) + x^(a - c)) = 1`
Simplify the following:
`(5^(n+3)-6xx5^(n+1))/(9xx5^x-2^2xx5^n)`
Solve the following equation for x:
`4^(2x)=1/32`
Find the value of x in the following:
`(root3 4)^(2x+1/2)=1/32`
If `3^(4x) = (81)^-1` and `10^(1/y)=0.0001,` find the value of ` 2^(-x+4y)`.
Solve the following equation:
`3^(x-1)xx5^(2y-3)=225`
Solve the following equation:
`sqrt(a/b)=(b/a)^(1-2x),` where a and b are distinct primes.
When simplified \[\left( - \frac{1}{27} \right)^{- 2/3}\] is
If \[2^{- m} \times \frac{1}{2^m} = \frac{1}{4},\] then \[\frac{1}{14}\left\{ ( 4^m )^{1/2} + \left( \frac{1}{5^m} \right)^{- 1} \right\}\] is equal to
The value of \[\sqrt{5 + 2\sqrt{6}}\] is
