Practicing Success
If $b + c = ax, c + a = by, a + b = cz$, then the value $\frac{1}{9}\left[\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right]$ is: |
$\frac{1}{9}$ 1 0 $\frac{1}{3}$ |
$\frac{1}{9}$ |
If $b + c = ax, c + a = by, a + b = cz$, then the value $\frac{1}{9}\left[\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right]$ Put the values of a = 1 , b = 1 and c = 1 Then by putting these values in the given equations the we will get the values of a = 2, b = 2 and c = 2 Now put these values in desired equation, $\frac{1}{9}\left[\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right]$ = $\frac{1}{9}\left[\frac{1}{2+1}+\frac{1}{2+1}+\frac{1}{2+1}\right]$ = $\frac{1}{9}\left[\frac{1}{3}+\frac{1}{3}+\frac{1}{3}\right]$ = \(\frac{1}{9}\) × \(\frac{3}{3}\) = \(\frac{1}{9}\) |