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
Find:-
`64^(1/2)`
Find:-
`9^(3/2)`
Prove that:
`(x^a/x^b)^(a^2+ab+b^2)xx(x^b/x^c)^(b^2+bc+c^2)xx(x^c/x^a)^(c^2+ca+a^2)=1`
Simplify:
`((5^-1xx7^2)/(5^2xx7^-4))^(7/2)xx((5^-2xx7^3)/(5^3xx7^-5))^(-5/2)`
Prove that:
`(2^(1/2)xx3^(1/3)xx4^(1/4))/(10^(-1/5)xx5^(3/5))div(3^(4/3)xx5^(-7/5))/(4^(-3/5)xx6)=10`
Find the value of x in the following:
`2^(5x)div2x=root5(2^20)`
State the power law of exponents.
If \[x = 7 + 4\sqrt{3}\] and xy =1, then \[\frac{1}{x^2} + \frac{1}{y^2} =\]
If \[x = \frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}}\] and \[y = \frac{\sqrt{5} - \sqrt{3}}{\sqrt{5} + \sqrt{3}}\] then x + y +xy=
Find:-
`32^(1/5)`
