em cảm ơn nhìuu
a: \(x\left(x-1\right)-\left(x+2\right)^2=x+8\)
=>\(x^2-x-x^2-4x-4=x+8\)
=>-5x-4=x+8
=>-5x-x=4+8
=>-6x=12
=>x=-2
b: \(4x^2\left(x-1\right)+9-9x=0\)
=>\(4x^2\left(x-1\right)-9\left(x-1\right)=0\)
=>\(\left(x-1\right)\left(4x^2-9\right)=0\)
=>\(\left(x-1\right)\left(2x-3\right)\left(2x+3\right)=0\)
=>\(\left[{}\begin{matrix}x-1=0\\2x-3=0\\2x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
c: \(x^3=2x^2-x\)
=>\(x^3-2x^2+x=0\)
=>\(x\left(x^2-2x+1\right)=0\)
=>\(x\left(x-1\right)^2=0\)
=>\(\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
d: \(2x^2-7x+3=0\)
=>\(2x^2-6x-x+3=0\)
=>2x(x-3)-(x-3)=0
=>(x-3)(2x-1)=0
=>\(\left[{}\begin{matrix}x-3=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{2}\end{matrix}\right.\)
