Đáp án:a/ x=45; y=60; z=84
b/
$\begin{array}{l}
x = \frac{1}{2}\\
y = \frac{5}{6}\\
z = - \frac{5}{6}
\end{array}$
Giải thích các bước giải:a/
$\begin{array}{l}
\frac{x}{3} = \frac{y}{4} = > \frac{x}{{15}} = \frac{y}{{20}}\\
\frac{y}{5} = \frac{z}{7} = > \frac{y}{{20}} = \frac{z}{{28}}\\
= > \frac{x}{{15}} = \frac{y}{{20}} = \frac{z}{{28}} = \frac{{2x + 3y - z}}{{2.15 + 3.20 - 28}} = \frac{{186}}{{62}} = 3\\
= > x = 3.15 = 45\\
= > y = 3.20 = 60\\
= > z = 3.28 = 84
\end{array}$
b/
$\begin{array}{l}
\frac{{y + z + 1}}{x} = \frac{{x + z + 2}}{y} = \frac{{x + y - 3}}{z} = \frac{1}{{x + y + z}} = \frac{{(y + z + 1) + (x + z + 2) + (x + y - 3)}}{{x + y + z}} = \frac{{2(x + y + z)}}{{(x + y + z)}} = 2\\
= > 2(x + y + z) = 1 = > x + y + z = \frac{1}{2}\\
lại có \\
y + z + 1 = 2x = > x + y + z = 3x - 1 < = > \frac{1}{2} = 3x - 1 = > x = \frac{1}{2}\\
x + z + 2 = 2y = > x + y + z = 3y - 2 < = > \frac{1}{2} = 3y - 2 = > y = \frac{5}{6}\\
x + y - 3 = 2z = > x + y + z = 3z + 3 < = > \frac{1}{2} = 3z + 3 = > z = - \frac{5}{6}
\end{array}$
