\(3\left(2x-3\right)+2\left(2-x\right)=3\)
=> 6x - 9 + 4 - 2x = 3
=> 4x - 5 = 3
=> 4x = 8
=> x = 2
\(3\left(2x-3\right)+2\left(2-x\right)=-3\)
\(\Leftrightarrow6x-9+4-2x=-3\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
3.( 2x - 3) + 2(2 - x) = -3
⇒ 6x - 9 + 4 - 2x = -3
⇒ 4x - 5 = -3
⇒ 4x = 2
⇒ x = \(\dfrac{1}{2}\)
\(3\left(2x-3\right)+2\left(2-x\right)=-3\)
⇔ \(6x-9+4-2x=-3\)
⇔ \(4x-9+4=-3\)
⇔ \(4x-5=-3\)
⇔ \(4x=2\)
⇔ \(x=\dfrac{1}{2}\)
Chị gửi em nha,đúng tích cho chị nha ^^