Đáp án:
$4)\\ a) x =-1\\ b) \left[\begin{array}{l} x=0 \\ x =1\end{array} \right.\\ c) \left[\begin{array}{l} x=0 \\ x=-1\\x=1\end{array} \right.\\ d)x=2$
Giải thích các bước giải:
$4)\\ a)x^2(x+1)+x+1=0\\ \Leftrightarrow (x^2+1)(x+1)=0\\ \Leftrightarrow \left[\begin{array}{l} x^2+1=0 \\ x+1 =0\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} x^2=-1(\text{Vô nghiệm}) \\ x =-1\end{array} \right.\\ \Leftrightarrow x =-1\\ b)x^2-x=-2x^2+2x\\ \Leftrightarrow x^2-x+2x^2-2x=0\\ \Leftrightarrow 3x^2-3x=0\\ \Leftrightarrow 3x(x-1)=0\\ \Leftrightarrow \left[\begin{array}{l} x=0 \\ x-1 =0\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} x=0 \\ x =1\end{array} \right.\\ c)x^2(x-1)+x^2=x\\ \Leftrightarrow x^2(x-1)+x^2-x=0\\ \Leftrightarrow x^2(x-1)+x(x-1)=0\\ \Leftrightarrow (x^2+x)(x-1)=0\\ \Leftrightarrow x(x+1)(x-1)=0\\ \Leftrightarrow \left[\begin{array}{l} x=0 \\ x+1=0\\x-1=0\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} x=0 \\ x=-1\\x=1\end{array} \right.\\ d)(x-2)(x^2+4)=x^2-2x\\ \Leftrightarrow (x-2)(x^2+4)-x^2+2x=0\\ \Leftrightarrow (x-2)(x^2+4)-x(x-2)=0\\ \Leftrightarrow (x-2)(x^2-x+4)=0\\ \Leftrightarrow (x-2)\left(x^2-x+\dfrac{1}{4}+\dfrac{15}{4}\right)=0\\ \Leftrightarrow (x-2)\left(\left(x-\dfrac{1}{2}\right)^2+\dfrac{15}{4}\right)=0\\ \Leftrightarrow x-2=0\left(\text{Do }\left(x-\dfrac{1}{2}\right)^2+\dfrac{15}{4}>0 \ \forall \ x\right)\\ \Leftrightarrow x=2$
