a/ Ta có ;
\(x+y+z=92\)
\(\dfrac{x}{2}=\dfrac{y}{3};\dfrac{y}{5}=\dfrac{z}{7}\)
\(\Leftrightarrow\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{21}\)
Áp dụng t/c dãy tỉ số bằng nhau ta có :
\(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{21}=\dfrac{x+y+z}{10+15+21}=\dfrac{92}{46}=2\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{10}=2\Leftrightarrow x=20\\\dfrac{y}{15}=2\Leftrightarrow y=30\\\dfrac{z}{21}=2\Leftrightarrow z=42\end{matrix}\right.\)
Vậy .................
b/Ta có :
\(x+y-z=95\)
\(2x=3y=5z\)
\(\Leftrightarrow\dfrac{2x}{30}=\dfrac{3y}{30}=\dfrac{5z}{30}\)
\(\Leftrightarrow\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{5}\)
Áp dụng t/x dãy tỉ số bằng nhau ta có :
\(\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{5}=\dfrac{x+y-z}{15+10-5}=\dfrac{95}{19}=5\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{15}=5\Leftrightarrow x=75\\\dfrac{y}{10}=5\Leftrightarrow y=50\\\dfrac{z}{5}=5\Leftrightarrow z=25\end{matrix}\right.\)
Vậy ..
a, \(\dfrac{x}{2}=\dfrac{y}{3},\dfrac{y}{5}=\dfrac{z}{7},x+y+z=92\)
Ta có: \(\dfrac{x}{2}=\dfrac{y}{3}\Leftrightarrow\dfrac{x}{10}=\dfrac{y}{15}\left(1\right)\)
\(\dfrac{y}{5}=\dfrac{z}{7}\Leftrightarrow\dfrac{y}{15}=\dfrac{z}{21}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{21},x+y+z=92\)
AD t/c DTS = nhau ta có:
\(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{21}=\dfrac{x+y+z}{10+15+21}=\dfrac{92}{46}=2\)
+) \(\dfrac{x}{10}=2\Rightarrow x=20\)
+) \(\dfrac{y}{15}=2\Rightarrow y=30\)
+) \(\dfrac{z}{21}=2\Rightarrow z=42\)
b, \(2x=3y=5z,x+y-z=95\)
\(\Rightarrow\dfrac{30x}{15}=\dfrac{30y}{10}=\dfrac{30z}{6}\Rightarrow\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6},x+y-z=95\)
AD t/c DTS = nhau ta có:
\(\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}=\dfrac{x+y-z}{15+10-6}=\dfrac{95}{19}=5\)
+) \(\dfrac{x}{15}=5\Rightarrow x=75\)
+) \(\dfrac{y}{10}=5\Rightarrow y=50\)
+) \(\dfrac{z}{6}=5\Rightarrow z=30\)
c, Bn xem lại đề bài nha! 
