Question

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

Answer 1

Sửa đề: \(\dfrac{7-2x}{4}+\dfrac{3x+2}{6}=\dfrac{3-x}{3}\)

Ta có: \(\dfrac{7-2x}{4}+\dfrac{3x+2}{6}=\dfrac{3-x}{3}\)

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

\(\Leftrightarrow21-6x+6x+4=12-4x\)

\(\Leftrightarrow-4x+12=25\)

\(\Leftrightarrow-4x=13\)

\(\Leftrightarrow x=-\dfrac{13}{4}\)

Vậy: \(S=\left\{-\dfrac{13}{4}\right\}\)

Answer 2

7-2x/4+3x+2/6=3-x/3

⇔6(7-2x)/24+4(3x+2)/24=8(3-x)/24

⇔42-12x+12x+8=24-8x

⇔-12x+12x+8x=24-42-8

⇔8x=-26

⇔x=-13/4

phương trình có 1 nghiệm x=-13/4