a: (x-2)(x+2)-(x+1)2=1
=>\(x^2-4-\left(x^2+2x+1\right)=1\)
=>\(x^2-4-x^2-2x-1=1\)
=>-2x-5=1
=>-2x=6
=>\(x=\dfrac{6}{-2}=-3\)
b: Sửa đề:\(x^3-8-\left(x-2\right)\left(x-4\right)=0\)
=>\(\left(x^3-8\right)-\left(x-2\right)\left(x-4\right)=0\)
=>\(\left(x-2\right)\left(x^2+2x+4\right)-\left(x-2\right)\left(x-4\right)=0\)
=>\(\left(x-2\right)\left(x^2+2x+4-x+4\right)=0\)
=>\(\left(x-2\right)\left(x^2+x\right)=0\)
=>x(x+1)(x-2)=0
=>\(\left[{}\begin{matrix}x=0\\x+1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=2\end{matrix}\right.\)
c: 3x(x-1)+1-x=0
=>3x(x-1)-(x-1)=0
=>(x-1)(3x-1)=0
=>\(\left[{}\begin{matrix}x-1=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)
