#LC
a, `x^3 - 2x^2 + x=0` ⇔ `x(x^2-2x+1) = 0` ⇔ `x(x-1)^2 = 0`
⇔\(\left[ \begin{array}{l}x=0\\x=1\end{array} \right.\)
b, `-5x^2+30x=0` ⇔ `5x(6-x) = 0`
⇔\(\left[ \begin{array}{l}x=0\\x=6\end{array} \right.\)
c, `3x(x-2020)- x+2020=0` ⇔ `3x(x-2020) - (x-2020)=0` ⇔ `(x-2020)(3x-1)=0`
⇔\(\left[ \begin{array}{l}x=\frac{1}{3}\\x=2020\end{array} \right.\)
d, `(x^2-4)-(x-2)(3-2x)=0` ⇔ `(x-2)(x+2) - (x-2)(3-2x) = 0`
⇔ `(x-2)(x+2-3+2x)=0` ⇔ `(x-2)(3x-1)=0`
⇔\(\left[ \begin{array}{l}x=2\\x=\frac{1}{3}\end{array} \right.\)
e, `2x(x-3) + 5(x-3)=0` ⇔ `(x-3)(2x+5)=0`
⇔\(\left[ \begin{array}{l}x=3\\x=\frac{-5}{2}\end{array} \right.\)
f, `x^3+27+(x+3)(x-9)=0`⇔ `(x^3+ 3^3) +(x+3)(x-9) = 0`
⇔ `(x+3)(x^2-3x+9) + (x-9)(x+3) = 0` ⇔ `(x+3)(x^2-3x+9+x-9) = 0`
⇔ `(x+3)(x^2-2x)=0` ⇔ `x(x+3)(x-2)=0`
⇔\(\left[ \begin{array}{l}x=2\\x=0\\x=-3\end{array} \right.\)
g, `(2x+1)(4x-3) = (5x-2)(2x+1)` ⇔ `(2x+1)(4x-3) - (5x-2)(2x+1)=0`
⇔ `(2x+1)(4x-3-5x+2)=0` ⇔ `(2x+1)(-x-1) = 0`
⇔\(\left[ \begin{array}{l}x=\frac{-1}{2}\\x=-1\end{array} \right.\)
h, `x^2 (2x-1)=6x-3` ⇔ `x^2(2x-1) - (6x-3)=0`
⇔`x^2(2x-1) - 3(2x-1)=0` ⇔ `(2x-1)(x^2-3) = 0`
⇔ \(\left[ \begin{array}{l}x=\frac{1}{2}\\x^{2}=3\end{array} \right.\)
⇔⇔ \(\left[ \begin{array}{l}x=\frac{1}{2}\\x=\pm\sqrt{3}\end{array} \right.\)
i, `(x-2)(x^2+3x-2)-x^3=-8` ⇔ `(x-2)(x^2+3x-2)-(x^3-2^3)=0`
⇔`(x-2)(x^2+3x-2) - (x-2)(x^2+2x+4)=0`
⇔ `(x-2)(x^2+3x-2-x^2-2x-4)=0` ⇔ `(x-2)(x-6)=0`
⇔\(\left[ \begin{array}{l}x=2\\x=6\end{array} \right.\)
