REAL NUMBERS

if,
2x=3y=6-z,then what is the value of
1x+1y+1z..........?????
plz do reply fast........

5 Answers

269
Astha Gupta ·

any one...??????

2305
Shaswata Roy ·

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}}}

269
Astha Gupta ·

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....

2305
Shaswata Roy ·

\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}

269
Astha Gupta ·

oh...ya....u r right....dat was 1 not 6..........n d rest of d part i understood....thanks a lot.....

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