a) $3^x+3^{x+1}+3^{x+2}=351$
=> $3^{x}$ (1 + 3 + 9) = 351
=> $3^{x}$ . 13 = 351=> $3^{x}$ = 351 : 13 = 27
=> x = 3
b)
$\frac{x}{z+y+1}$ = $\frac{y}{x + z +1}$ = $\frac{z}{x+y-2}$ = x + y + z
= $\frac{x+y+z}{z+y+1+x+z+1+x+y-2}$
= $\frac{x+y+z}{2x+2y+2z}$
= $\frac{x+y+z}{2(x+y+z)}$
= $\frac{1}{2}$
*$\frac{x}{z+y+1}=\frac{1}{2}\Rightarrow 2x=z+y+1\Rightarrow z+y=2x-1$ (1)
Lại có: x + y + z = $\frac{1}{2}$ (2)
Thay (1) vào (2) ta đc:
x + 2x - 1 = $\frac{1}{2}$
=> 3x = $\frac{1}{2}$ + 1 = $\frac{3}{2}$
=> x = $\frac{3}{2}$ : 3 = $\frac{1}{2}$
*$\frac{y}{x+z+1}=\frac{1}{2}\Rightarrow 2y=x+z+1\Rightarrow x+z=2y-1$ (3)
Thay (3) vào (2) ta đc:
2y - 1 + y = $\frac{1}{2}$
=> 3y = $\frac{1}{2}$ +1 = $\frac{3}{2}$
=> y = $\frac{3}{2}$ : 3 = $\frac{1}{2}$
*x + y + z = $\frac{1}{2}$
=> z = $\frac{1}{2}$ - x - y = $\frac{1}{2}$ - $\frac{1}{2}$ - $\frac{1}{2}$ = $\frac{-1}{2}$
Vậy:.............