\(\frac{x}{\frac{1}{2}}=\frac{y}{\frac{1}{3}}=\frac{z}{\frac{1}{5}}\)
vì \(\left|x+y-z\right|=95\Rightarrow\orbr{\begin{cases}x+y-z=95\\x+y-z=-95\end{cases}}\)
th1: x+y-z=95
áp dụng t/c dãy tỉ số bằng nhau ta có:
\(\frac{x}{\frac{1}{2}}=\frac{y}{\frac{1}{3}}=\frac{z}{\frac{1}{5}}=\frac{x+y-z}{\frac{1}{2}+\frac{1}{3}-\frac{1}{5}}=\frac{95}{\frac{19}{30}}=150\)
\(\frac{x}{\frac{1}{2}}=150\Rightarrow x=75\)
\(\frac{y}{\frac{1}{3}}=150\Rightarrow y=50\)
\(\frac{z}{\frac{1}{5}}=150\Rightarrow z=30\)
th2: x+y-z=-95
\(\frac{x}{\frac{1}{2}}=\frac{y}{\frac{1}{3}}=\frac{z}{\frac{1}{5}}=\frac{x+y-z}{\frac{1}{2}+\frac{1}{3}-\frac{1}{5}}=-\frac{95}{\frac{19}{30}}=-150\)
\(\frac{x}{\frac{1}{2}}=-150\Rightarrow x=-75\)
\(\frac{y}{\frac{1}{3}}=-150\Rightarrow y=-50\)
\(\frac{z}{\frac{1}{5}}=-150\Rightarrow z=-30\)
vậy x=75, y=50,z=30
hay x=-75, y=-50, x=-30
