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 \(x=\dfrac{1}{2}\)
