Let 2x=3y=6-z=k
\large \mathbf{2=k^{\frac{1}{x}},3=k^{\frac{1}{y}},6=k^{\frac{-1}{z}}}
Now try to find \large \mathbf{k^{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}}
if,
2x=3y=6-z,then what is the value of
1x+1y+1z..........?????
plz do reply fast........
Let 2x=3y=6-z=k
\large \mathbf{2=k^{\frac{1}{x}},3=k^{\frac{1}{y}},6=k^{\frac{-1}{z}}}
Now try to find \large \mathbf{k^{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}}
i think...it must b smthng lyk..
k1x x kiy x k-1z
=2x3x16
=6
i dont knw whether this is right or not.....????
kindly check it....
\large \mathbf{k^{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}=\frac{2\times 3}{6}=1=k^{0}}
\rightarrow \mathbf{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0}
oh...ya....u r right....dat was 1 not 6..........n d rest of d part i understood....thanks a lot.....