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
Assuming that x, y, z are positive real numbers, simplify the following:
`sqrt(x^3y^-2)`
Assuming that x, y, z are positive real numbers, simplify the following:
`root5(243x^10y^5z^10)`
Show that:
`{(x^(a-a^-1))^(1/(a-1))}^(a/(a+1))=x`
Show that:
`(3^a/3^b)^(a+b)(3^b/3^c)^(b+c)(3^c/3^a)^(c+a)=1`
Solve the following equation:
`4^(x-1)xx(0.5)^(3-2x)=(1/8)^x`
If 24 × 42 =16x, then find the value of x.
Write the value of \[\sqrt[3]{7} \times \sqrt[3]{49} .\]
The value of x − yx-y when x = 2 and y = −2 is
Which one of the following is not equal to \[\left( \sqrt[3]{8} \right)^{- 1/2} ?\]
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
