Ta có: \(\dfrac{2x-1}{2}=\dfrac{x+3}{4}+\dfrac{x+2}{2}\)
\(\Leftrightarrow\dfrac{2\left(2x-1\right)}{4}=\dfrac{x+3}{4}+\dfrac{2\left(x+2\right)}{4}\)
\(\Leftrightarrow4x-2=x+3+2x+4\)
\(\Leftrightarrow4x-2=3x+7\)
\(\Leftrightarrow4x-3x=7+2\)
hay x=9
Vậy: x=9
`(2x-1)/2=(x+3)/4+(x+2)/2`
`<=>2(2x-1)=x+3+2(x+2)`
`<=>4x-2=x+3+2x+4`
`<=>4x-2=3x+7`
`<=>x=9`
Vậy `x=9`
\(\dfrac{2x-1}{2}=\dfrac{x+3}{4}+\dfrac{x+2}{2}\)
\(\Rightarrow\dfrac{2\left(2x-1\right)}{4}-\dfrac{x+3}{4}-\dfrac{2\left(x+2\right)}{4}=0\)
\(\Rightarrow\dfrac{4x-2-x-3-2x-4}{4}=0\)
\(\Rightarrow x-9=0\)
=> x = 9
Vây x = 9
\(\dfrac{2x-1}{2}=\dfrac{x+3}{4}+\dfrac{x+2}{2}\)
\(\dfrac{4x-2}{4}=\dfrac{x+3}{4}+\dfrac{2x+4}{4}\)
\(\Rightarrow4x-2=x+3+2x+4\)
\(2x=9\)
\(x=\dfrac{9}{2}\)
