Advertisements
Advertisements
प्रश्न
If x = 2 and y = 4, then \[\left( \frac{x}{y} \right)^{x - y} + \left( \frac{y}{x} \right)^{y - x} =\]
पर्याय
4
8
12
2
Advertisements
उत्तर
We have to find the value of `(x/y)^(x-y) + (y/x)^(y-x) ` if x = 2, y = 4
Substitute x = 2, y = 4 in `(x/y)^(x-y) + (y/x)^(y-x) ` to get,
`(x/y)^(x-y) + (y/x)^(y-x) ` = `(2/4)^(2-4) + (4/2)^(4-2)`
= `(2/4)^-2+ (4/2)^2`
= `(1/2)^-2 + (2)^2`
= `(1/2^-2) + 4`
`(x/y)^(x-y) + (y/x)^(y-x) = 1/(1/2^2) +4`
=` 1/(1/4) +4`
= `1 xx 4/1 +4`
= 4+4
= 8
APPEARS IN
संबंधित प्रश्न
Prove that:
`(a+b+c)/(a^-1b^-1+b^-1c^-1+c^-1a^-1)=abc`
Prove that:
`(a^-1+b^-1)^-1=(ab)/(a+b)`
Assuming that x, y, z are positive real numbers, simplify the following:
`sqrt(x^3y^-2)`
Show that:
`(x^(1/(a-b)))^(1/(a-c))(x^(1/(b-c)))^(1/(b-a))(x^(1/(c-a)))^(1/(c-b))=1`
Find the value of x in the following:
`2^(5x)div2x=root5(2^20)`
Find the value of x in the following:
`2^(x-7)xx5^(x-4)=1250`
Solve the following equation:
`4^(2x)=(root3 16)^(-6/y)=(sqrt8)^2`
If \[\frac{3^{2x - 8}}{225} = \frac{5^3}{5^x},\] then x =
If x = \[\frac{2}{3 + \sqrt{7}}\],then (x−3)2 =
\[\frac{1}{\sqrt{9} - \sqrt{8}}\] is equal to
