Advertisements
Advertisements
प्रश्न
If `x = a^(m + n), y = a^(n + l)` and `z = a^(l + m),` prove that `x^my^nz^l = x^ny^lz^m`
Advertisements
उत्तर
Given `x = a^(m + n), y = a^(n + l)` and `z = a^(l + m)`
Putting the values of x, y and z in `x^my^nz^l,` we get
`x^my^nz^l`
= `(a^(m + n))^m(a^(n + l))^n(a^(l + m))^l`
= `(a^(m^2 + nm))(a^(n^2 + ln))(a^(l^2 + lm))`
= `a^(m^2 + n^2 + l^2 + nm + ln + lm)`
Putting the values of x, y and z in `x^ny^lz^m,` we get
`x^ny^lz^m`
= `(a^(m + n))^n(a^(n + l))^l(a^(l + m))^m`
= `(a^(mn + n^2))(a^(nl + l^2))(a^(lm + m^2))`
= `a^(mn+n^2 + nl + l^2 + lm + m^2)`
So, `x^my^nz^l = x^ny^lz^m`.
APPEARS IN
संबंधित प्रश्न
Find:-
`64^(1/2)`
Show that:
`{(x^(a-a^-1))^(1/(a-1))}^(a/(a+1))=x`
Solve the following equation:
`4^(x-1)xx(0.5)^(3-2x)=(1/8)^x`
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.
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 is a positive real number and x2 = 2, then x3 =
The value of \[\frac{\sqrt{48} + \sqrt{32}}{\sqrt{27} + \sqrt{18}}\] is
If \[\sqrt{2} = 1 . 4142\] then \[\sqrt{\frac{\sqrt{2} - 1}{\sqrt{2} + 1}}\] is equal to
Find:-
`125^(1/3)`
If `a = 2 + sqrt(3)`, then find the value of `a - 1/a`.
