Giải:
\(16x^2-9\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(4x\right)^2-\left[3\left(x+1\right)\right]^2=0\)
\(\Leftrightarrow\left(4x\right)^2-\left(3x+3\right)^2=0\)
\(\Leftrightarrow\left[4x-\left(3x+3\right)\right]\left[4x+\left(3x+3\right)\right]=0\)
\(\Leftrightarrow\left(4x-3x-3\right)\left(4x+3x+3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(7x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\7x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0+3\\7x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{3}{7}\end{matrix}\right.\)
Vậy \(x=3\) hoặc \(x=-\dfrac{3}{7}\).
Chúc bạn học tốt!