Bổ sung bài 2 :

a ) $x^2\left(x+1\right)+2x\left(x+1\right)=0$

$\Leftrightarrow\left(x+1\right)\left(x^2+2x\right)=0$

$\Leftrightarrow x\left(x+1\right)\left(x+2\right)=0$

$\Leftrightarrow\left\{{}\begin{matrix}x=0\\x+1=0\\x+2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\\x=-1\\x=-2\end{matrix}\right.$

Vậy $x=0;x=-1;x=-2$

b ) $\dfrac{4}{9}-25x^2=0$

$\Leftrightarrow\left(\dfrac{2}{3}\right)^2-\left(5x\right)^2=0$

$\Leftrightarrow\left(\dfrac{2}{3}-5x\right)\left(\dfrac{2}{3}+5x\right)=0$

$\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{3}-5x=0\\\dfrac{2}{3}+5x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{2}{15}\\x=-\dfrac{2}{15}\end{matrix}\right.$

Vậy $x=\dfrac{2}{15};x=-\dfrac{2}{15}$

c ) $x\left(3x-2\right)-5\left(2-3x\right)=0$

$\Leftrightarrow x\left(3x-2\right)+5\left(3x-2\right)=0$

$\Leftrightarrow\left(3x-2\right)\left(x+5\right)=0$

$\Leftrightarrow\left\{{}\begin{matrix}3x-2=0\\x+5=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\x=-5\end{matrix}\right.$

Vậy $x=\dfrac{2}{3}$ và $x=-5$

d ) $x^2-x+\dfrac{1}{4}=0$

$\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2=0$

$\Rightarrow x=\dfrac{1}{2}$

Vậy $x=\dfrac{1}{2}$