Giải:
\(\dfrac{7-2x}{5}=\dfrac{3-x}{2}\)
\(\Leftrightarrow2\left(7-2x\right)=5\left(3-x\right)\)
\(\Leftrightarrow14-4x=15-5x\)
\(\Leftrightarrow-4x+5x=15-14\)
\(\Leftrightarrow x=1\)
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Ta có: \(\dfrac{7-2x}{5}=\dfrac{3-x}{2}\)
\(\Rightarrow\) \(\left(7-2x\right).2=5.\left(3-x\right)\)
\(\Rightarrow\) \(14-4x=15-5x\)
\(\Rightarrow\) \(15-14=5x+\left(-4x\right)\)
\(\Rightarrow x=1\)
Vậy \(x=1\)
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