a: \(x^2-x=-2x+2\)
=>\(x\left(x-1\right)+2x-2=0\)
=>x(x-1)+2(x-1)=0
=>(x-1)(x+2)=0
=>\(\left[{}\begin{matrix}x-1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
b: \(4x^2+4x+1=x^2\)
=>\(3x^2+4x+1=0\)
=>(x+1)(3x+1)=0
=>\(\left[{}\begin{matrix}x+1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{1}{3}\end{matrix}\right.\)
c: \(x^2+\left(x+2\right)\left(11x-7\right)=9\)
=>\(x^2+11x^2-7x+22x-14-9=0\)
=>\(12x^2+15x-23=0\)
\(\Delta=15^2-4\cdot12\cdot\left(-23\right)=1329>0\)
=>Phương trình có hai nghiệm phân biệt là:
\(\left[{}\begin{matrix}x=\dfrac{-15-\sqrt{1329}}{2\cdot12}=\dfrac{-15-\sqrt{1329}}{24}\\x=\dfrac{-15+\sqrt{1329}}{24}\end{matrix}\right.\)
d: \(5x^3+4x=0\)
=>\(x\left(5x^2+4\right)=0\)
mà \(5x^2+4>=4>0\forall x\)
nên x=0
e: \(18x-9x^2=0\)
=>-9x(x-2)=0
=>x(x-2)=0
=>\(\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
