Question

giải phương trình sau : 

\(\left(2x^2-x\right)-9\left(2x^2-x\right)+18=0\) (đặt \(a=2x^2-x\))

Answer 1

Đề là \(\left(2x^2-x\right)^2+...\) hay là \(\left(2x^2-x\right)+...\) vậy bn?

Answer 2

Đặt \(2x^2-x=a\)

\(PT\Leftrightarrow a^2-9a+18=0\\ \Leftrightarrow a^2-3a-6a+18=0\\ \Leftrightarrow\left(a-3\right)\left(a-6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x^2-x-3=0\\2x^2-x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left(2x-3\right)\left(x+1\right)=0\\\left(x-2\right)\left(2x+3\right)=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-1\\x=2\\x=-\dfrac{3}{2}\end{matrix}\right.\)

Answer 3

\(\left(2x^2-x\right)+9\left(2x^2-x\right)+18=0\)

⇔\(\left(2x^2-x\right)\left(1+9\right)+18=0\)

⇔\(10\left(2x^2-x\right)+18=0\)

⇔\(10\left(2x^2-x\right)=-18\)

⇔\(2x^2-x=-\dfrac{9}{5}\)

⇔\(x\left(2x-1\right)=-\dfrac{9}{5}\)

⇔\(x=-\dfrac{9}{5}\) hay \(2x-1=-\dfrac{9}{5}\)

⇔\(x=-\dfrac{9}{5}\) hay \(2x=-\dfrac{4}{5}\)

⇔\(x=-\dfrac{9}{5}\) hay \(x=-\dfrac{2}{5}\)