Ta có:
\(\dfrac{1}{3}x+2=\dfrac{2x-1}{2}\)
\(\Leftrightarrow\dfrac{1}{3}x+2=\dfrac{2x}{2}-\dfrac{1}{2}\)
\(\Leftrightarrow\) \(\dfrac{1}{3}x+2=x-\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{1}{3}x=x-\left(\dfrac{1}{2}+2\right)\)
\(\Leftrightarrow\dfrac{1}{3}x=x-\dfrac{5}{2}\)
\(\Leftrightarrow\)\(x-\dfrac{1}{3}x=\dfrac{5}{2}\)
\(\Leftrightarrow x\left(1-\dfrac{1}{3}\right)=\dfrac{5}{2}\)
\(\Leftrightarrow\dfrac{2}{3}x=\dfrac{5}{2}\)
\(\Leftrightarrow\) \(x=\dfrac{5}{2}:\dfrac{2}{3}=\dfrac{15}{4}\)
Vậy: x=\(\dfrac{15}{4}\)
( có thể đề bài như trên hoặc: \(\dfrac{1}{3}x+2=\)2x-\(\dfrac{1}{2}\))
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