Advertisements
Advertisements
प्रश्न
If 2x = 3y = 6-z, show that `1/x+1/y+1/z=0`
Advertisements
उत्तर
Let 2x = 3y = 6-z = k
`rArr2=k^(1/x),` `3=k^(1/y),` `6=k^(1/-z)`
Now,
`6 = 2xx3=k^(1/-z)`
`rArrk^(1/x)xxk^(1/y)=k^(1/-z)`
`rArrk^(1/x+1/y)=k^(1/-z)`
`rArr1/x+1/y=1/-z`
`rArr1/x+1/y+1/z=0`
APPEARS IN
संबंधित प्रश्न
Simplify:-
`2^(2/3). 2^(1/5)`
Simplify the following
`(4ab^2(-5ab^3))/(10a^2b^2)`
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`
Solve the following equations for x:
`2^(2x)-2^(x+3)+2^4=0`
Assuming that x, y, z are positive real numbers, simplify the following:
`(x^((-2)/3)y^((-1)/2))^2`
Prove that:
`9^(3/2)-3xx5^0-(1/81)^(-1/2)=15`
If 2x = 3y = 12z, show that `1/z=1/y+2/x`
Write the value of \[\sqrt[3]{7} \times \sqrt[3]{49} .\]
If \[\frac{2^{m + n}}{2^{n - m}} = 16\], \[\frac{3^p}{3^n} = 81\] and \[a = 2^{1/10}\],than \[\frac{a^{2m + n - p}}{( a^{m - 2n + 2p} )^{- 1}} =\]
The simplest rationalising factor of \[\sqrt{3} + \sqrt{5}\] is ______.
