Question

Giải phương trình:

\(\dfrac{2x-1}{4}+\dfrac{x-3}{3}=\dfrac{4x-2}{3}-\dfrac{6x+7}{12}\)

Answer 1

\(\Leftrightarrow\dfrac{6x-3}{12}+\dfrac{4x-12}{12}=\dfrac{16x-8}{12}-\dfrac{6x+7}{12}\)

\(\Leftrightarrow6x-3+4x-12=16x-8-6x-7\)

\(\Leftrightarrow0x=0\)

\(\Rightarrow\) phương trình vô số nghiệm

Answer 2

\(\dfrac{2x-1}{4}+\dfrac{x-3}{3}=\dfrac{4x-2}{3}-\dfrac{6x+7}{12}\) \(\Leftrightarrow\dfrac{2x-1}{4}+\dfrac{6x+7}{12}=\dfrac{4x-2}{3}-\dfrac{x-3}{3}\)

\(\Leftrightarrow\dfrac{3\left(2x-1\right)+6x+7}{12}=\dfrac{4\left(4x-2-x+3\right)}{12}\)

\(\Leftrightarrow\dfrac{6x-3+6x+7}{12}=\dfrac{16x-8-4x+12}{12}\)

\(\Leftrightarrow6x-3+6x+7=16x-8-4x+12\)

\(\Leftrightarrow4=12x+4\Leftrightarrow x=12x=0\)(phương trình vô nghiệm)

Vậy S=∅