Question

\(\dfrac{x-1}{2}-\dfrac{2x-1}{12}=\dfrac{2-x}{4}-\dfrac{3x+2}{6}\)

Answer 1

B1: Quy đồng 2 vế
+) MTC: 12
B2: Khử mẫu
B3: Giải phương trình
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Answer 2

\(\dfrac{x-1}{2}-\dfrac{2x-1}{12}=\dfrac{2-x}{4}-\dfrac{3x+2}{6}\\ \Leftrightarrow\dfrac{6\left(x-1\right)}{12}-\dfrac{2x-1}{12}-\dfrac{3\left(2-x\right)}{12}-\dfrac{2\left(3x+2\right)}{12}=0\\ \Leftrightarrow6x-6-2x+1-6+3x-6x-4=0\\ \Leftrightarrow x-15=0\\ \Leftrightarrow x=15\)

Answer 3

\(6x-6-2x+1=6-3x-6x-4\Leftrightarrow4x-5=2-9x\)

\(\Leftrightarrow13x=7\Leftrightarrow x=\dfrac{7}{13}\)

Answer 4

\(\Leftrightarrow\dfrac{6\left(x-1\right)}{12}-\dfrac{2x-1}{12}=\dfrac{3\left(2-x\right)}{12}-\dfrac{2\left(3x+2\right)}{12}\)

\(\Leftrightarrow6\left(x-1\right)-2x-1=3\left(2-x\right)-2\left(3x+2\right)\)

\(\Leftrightarrow6x-6-2x-1=6-3x-6x-4\)

\(\Leftrightarrow6x-2x+3x+6x=6-4+6+1\)

\(\Leftrightarrow13x=9\)

\(\Leftrightarrow x=\dfrac{9}{13}\)

Vậy: \(S=\left\{\dfrac{9}{13}\right\}\)