\(a,\Leftrightarrow x^2\left(x+5\right)-9\left(x+5\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\\x=-3\end{matrix}\right.\\ b,\Leftrightarrow\left[{}\begin{matrix}1-2x=3x-2\\2x-1=3x-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\\x=1\end{matrix}\right.\)
a) \(\Leftrightarrow x^2\left(x+5\right)-9x-45=0\)
\(\Leftrightarrow x^2\left(x+5\right)-9x\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x^2-9\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x-3=0\\x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=3\\x=-3\end{matrix}\right.\)
Vậy...
b) \(\Leftrightarrow\left(1-2x\right)^2-\left(3x-2\right)^2=0\)
\(\Leftrightarrow\left(1-2x-3x+2\right)\left(1-2x+3x-2\right)=0\)
\(\Leftrightarrow\left(3-5x\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3-5x=0\\x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\\x=1\end{matrix}\right.\)
Vậy...