\(x^2\left(x+1\right)+2x\left(x+1\right)=0\Leftrightarrow\left(x^2+2x\right)\left(x+1\right)=0\Leftrightarrow x\left(x+2\right)\left(x+1\right)=0\left\{{}\begin{matrix}x=0\\x+2=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=-2\\x=-1\end{matrix}\right.\)
a, \(x^2.\left(x+1\right)+2x\left(x+1\right)=0\)
=> ( x + 1 ) ( \(x^2\) + 2x ) = 0
=> ( x + 1 ) x (x + 2 ) = 0
=>\(\left[{}\begin{matrix}x=0\\x+1=0\\x+2=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=0\\x=-1\\x=-2\end{matrix}\right.\)
b, \(\dfrac{4}{9}-25x^2=0\)
=> \(\left(\dfrac{2}{3}\right)^2-\left(5x\right)^2=0\)
=> \(\left(\dfrac{2}{3}-5x\right)\left(\dfrac{2}{3}+5x\right)=0\)
=>\(\left[{}\begin{matrix}\dfrac{2}{3}-5x=0\\\dfrac{2}{3}+5x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{15}\\x=\dfrac{-2}{15}\end{matrix}\right.\)
b, \(\dfrac{4}{9}-25x^2=0\Leftrightarrow\left(5x\right)^2=-\dfrac{4}{9}\Leftrightarrow\left(5x\right)^2=-\left(\dfrac{2}{3}\right)^2\Leftrightarrow5x=-\dfrac{2}{3}\Leftrightarrow x=-\dfrac{2}{15}\)
a.\(x^2\left(x+1\right)+2x\left(x+1\right)=0\)
\(\left(x^2+2x\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+2x=0\\x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\left(x+2\right)=0\\x=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x+2=0\\x=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=-2\\x=-1\end{matrix}\right.\)
Vậy \(x=0;x=-2\) hoặc \(x=-1\) .
b.\(\dfrac{4}{9}-25x^2=0\)
\(\left(\dfrac{2}{3}\right)^2-\left(5x\right)^2=0\)
\(\left(\dfrac{2}{3}-5x\right)\left(\dfrac{2}{3}+5x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{3}-5x=0\\\dfrac{2}{3}+5x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{2}{3}\\5x=-\dfrac{2}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{15}\\x=-\dfrac{2}{15}\end{matrix}\right.\)
Vậy \(x=\dfrac{2}{15}\) hoặc \(x=-\dfrac{2}{15}\) .