Đáp án:
a) $x\in \left \{ 2;4 \right \}$
b) $x=0$
c) $x=\dfrac{1}{2}$
d) $x\in \left \{ \dfrac{1}{2020};\dfrac{1}{2021} \right \}$
Giải thích các bước giải:
a) $2(x-2)=x^{2}-4x+4$
$\leftrightarrow 2(x-2)=(x-2)^{2}$
$\leftrightarrow (x-2)^{2}-2(x-2)=0$
$\leftrightarrow (x-2)[(x-2)-2]=0$
$\leftrightarrow (x-2)(x-4)=0$
$\leftrightarrow \left[ \begin{array}{l}x-2=0\\x-4=0\end{array} \right.$
$\leftrightarrow \left[ \begin{array}{l}x=2\\x=4\end{array} \right.$
Vậy $x\in \left \{ 2;4 \right \}$
b) $(x+3)(x^{2}-3x+9)-x(x-2)^{2}=27$
$\leftrightarrow x^{3}+27-x(x^{2}-4x+4)=27$
$\leftrightarrow x^{3}+27-x^{3}+5x-4x=27$
$\leftrightarrow x+27=27$
$\leftrightarrow x=0$
Vậy $x=0$
c) $2x^{3}-x^{2}+2x-1=0$
$\leftrightarrow x^{2}(2x-1)+(2x-1)=0$
$\leftrightarrow (2x-1)(x^{2}+1)=0$
$\leftrightarrow \left[ \begin{array}{l}2x-1=0\\x^{2}+1=0\end{array} \right.$
$\leftrightarrow \left[ \begin{array}{l}x=\dfrac{1}{2}\\x\in \varnothing (Vì_{} x^{2}\geq 0\forall x)\end{array} \right.$
Vậy $x=\dfrac{1}{2}$
d) $2020x-1+2021x(1-2020x)=0$
$\leftrightarrow (2020x-1)-2021x(2020x-1)=0$
$\leftrightarrow (2020x-1)(1-2021x)=0$
$\leftrightarrow \left[ \begin{array}{l}2020x-1=0\\1-2021x=0\end{array} \right.$
$\leftrightarrow \left[ \begin{array}{l}x=\dfrac{1}{2020}\\x=\dfrac{1}{2021}\end{array} \right.$
Vậy $x\in \left \{ \dfrac{1}{2020};\dfrac{1}{2021} \right \}$
