Advertisements
Advertisements
प्रश्न
Assuming that x, y, z are positive real numbers, simplify the following:
`root5(243x^10y^5z^10)`
Advertisements
उत्तर
We have to simplify the following, assuming that x, y, z are positive real numbers
Given `root5(243x^10y^5z^10)`
`=(243xx x^10xxy^5xxz^10)^(1/5)`
`=(243)^(1/5)xx (x^10)^(1/5)xx(y^5)^(1/5)xx(z^10)^(1/5)`
`=(3^5)^(1/5)xx x^(10xx1/5)xxy^(5xx1/5)xxz^(10xx1/5)`
`=3xx x^2xxyxxz^2`
`=3x^2yz^2`
APPEARS IN
संबंधित प्रश्न
Solve the following equation for x:
`4^(x-1)xx(0.5)^(3-2x)=(1/8)^x`
Simplify:
`root3((343)^-2)`
Prove that:
`sqrt(1/4)+(0.01)^(-1/2)-(27)^(2/3)=3/2`
Show that:
`[{x^(a(a-b))/x^(a(a+b))}div{x^(b(b-a))/x^(b(b+a))}]^(a+b)=1`
Show that:
`(x^(a^2+b^2)/x^(ab))^(a+b)(x^(b^2+c^2)/x^(bc))^(b+c)(x^(c^2+a^2)/x^(ac))^(a+c)=x^(2(a^3+b^3+c^3))`
If `5^(3x)=125` and `10^y=0.001,` find x and y.
If (23)2 = 4x, then 3x =
If a, b, c are positive real numbers, then \[\sqrt{a^{- 1} b} \times \sqrt{b^{- 1} c} \times \sqrt{c^{- 1} a}\] is equal to
If \[x + \sqrt{15} = 4,\] then \[x + \frac{1}{x}\] =
If \[\sqrt{13 - a\sqrt{10}} = \sqrt{8} + \sqrt{5}, \text { then a } =\]
