1: =>-2x+6x2-4x+8=3/2+18x2
=>\(18x^2+\dfrac{3}{2}=6x^2-6x+8\)
\(\Leftrightarrow12x^2+6x-\dfrac{13}{2}=0\)
hay \(x\in\left\{\dfrac{-3+\sqrt{87}}{12};\dfrac{-3-\sqrt{87}}{12}\right\}\)
2: Đề thiếu vế phải rồi bạn
3: \(\Leftrightarrow x^2\cdot\dfrac{2}{3}-2x-\dfrac{4}{3}x+4=\dfrac{2}{3}x^2-\dfrac{2}{3}x\)
=>-10/3x+2/3x=-4
=>-8/3x=-4
=>x=4:8/3=4x3/8=12/8=3/2