Advertisements
Advertisements
प्रश्न
Show that:
`1/(1+x^(a-b))+1/(1+x^(b-a))=1`
Advertisements
उत्तर
`1/(1+x^(a-b))+1/(1+x^(b-a))`
`=1/(1+(x^a/x^b))+1/(1+(x^b/x^a))`
`=1/((x^b+x^a)/x^b)+1/((x^a+x^b)/x^a)`
`=x^b/(x^b+x^a)+x^a/(x^a+x^b)`
`=(x^b+x^a)/(x^b+x^a)`
= 1
APPEARS IN
संबंधित प्रश्न
Assuming that x, y, z are positive real numbers, simplify the following:
`(sqrt(x^-3))^5`
Prove that:
`sqrt(1/4)+(0.01)^(-1/2)-(27)^(2/3)=3/2`
Show that:
`(a^(x+1)/a^(y+1))^(x+y)(a^(y+2)/a^(z+2))^(y+z)(a^(z+3)/a^(x+3))^(z+x)=1`
Find the value of x in the following:
`5^(2x+3)=1`
If `a=x^(m+n)y^l, b=x^(n+l)y^m` and `c=x^(l+m)y^n,` Prove that `a^(m-n)b^(n-l)c^(l-m)=1`
If a, b, c are positive real numbers, then \[\sqrt[5]{3125 a^{10} b^5 c^{10}}\] is equal to
If 9x+2 = 240 + 9x, then x =
If \[\frac{x}{x^{1 . 5}} = 8 x^{- 1}\] and x > 0, then x =
\[\frac{1}{\sqrt{9} - \sqrt{8}}\] is equal to
If x = \[\sqrt[3]{2 + \sqrt{3}}\] , then \[x^3 + \frac{1}{x^3} =\]
