Lời giải ................
Bài 1 :
Câu a \(x^3-x^2-x+1=0\)
\(\Leftrightarrow\left(x^3-x^2\right)-\left(x-1\right)=0\)
\(\Leftrightarrow x^2\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-1\right)\left(x+1\right)\)
\(\Leftrightarrow\left(x-1\right)^2\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
Vậy \(x=1\) hoặc \(x=-1\)
Câu b : \(3\left(x-1\right)^2-3x\left(x-5\right)-2=0\)
\(\Leftrightarrow3x^2-6x+3-3x^2+15x-2=0\)
\(\Leftrightarrow9x+1=0\)
\(\Rightarrow x=-\dfrac{1}{9}\)
Vậy \(x=-\dfrac{1}{9}\)
Câu c : \(2x^2-5x-7=0\)
\(\Leftrightarrow2x^2+2x-7x-7=0\)
\(\Leftrightarrow\left(2x^2+2x\right)-\left(7x+7\right)=0\)
\(\Leftrightarrow2x\left(x+1\right)-7\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(2x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\2x-7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{7}{2}\end{matrix}\right.\)
Vậy \(x=-1\) hoặc \(x=\dfrac{7}{2}\)
