\(\dfrac{x}{3}=\dfrac{y}{2};\dfrac{x}{4}=\dfrac{z}{5}\) và \(x+y-z=10\)
Ta có:
\(\dfrac{x}{3}=\dfrac{y}{2}\Leftrightarrow\dfrac{x}{12}=\dfrac{y}{8};\dfrac{x}{4}=\dfrac{z}{5}\Leftrightarrow\dfrac{x}{12}=\dfrac{z}{15}\)
\(\Rightarrow\dfrac{y}{8}=\dfrac{x}{12}=\dfrac{z}{15}\) và \(x+y-z=10\)
AD tính chất DTS bằng nhau ta có:
\(\dfrac{y}{8}=\dfrac{x}{12}=\dfrac{z}{15}=\dfrac{x+y-z}{12+8-15}=\dfrac{10}{5}=2\)
+) \(\dfrac{y}{8}=2\Rightarrow y=16\)
+) \(\dfrac{x}{12}=2\Rightarrow x=42\)
+) \(\dfrac{z}{15}=2\Rightarrow z=30\)
Vậy \(x=42;y=16;z=30\)
c,\(\dfrac{x}{2}=\dfrac{y}{5};\dfrac{y}{3}=\dfrac{z}{2}\) và \(2x+3y-4z=34\)
Ta có:
\(\dfrac{x}{2}=\dfrac{y}{5}\Leftrightarrow\dfrac{x}{6}=\dfrac{y}{15};\dfrac{y}{3}=\dfrac{z}{2}\Leftrightarrow\dfrac{y}{15}=\dfrac{z}{10}\)
\(\Rightarrow\dfrac{x}{6}=\dfrac{y}{15}=\dfrac{z}{10}\)
Ta lại có:
\(\dfrac{2x}{12}=\dfrac{3y}{45}=\dfrac{4z}{40}\) và \(2x+3y-4z=34\)
AD tính chất DTS bằng nhau ta có:
\(\dfrac{2x}{12}=\dfrac{3y}{45}=\dfrac{4z}{40}=\dfrac{2x+3y-4z}{12+45-40}=\dfrac{34}{17}=2\)
+) \(\dfrac{2x}{12}=2\Rightarrow x=12\)
+) \(\dfrac{3y}{45}=2\Rightarrow y=30\)
+) \(\dfrac{4z}{40}=2\Rightarrow z=20\)
Vậy \(x=12;y=30;z=20\)
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