Giải:
ĐKXĐ: \(x\ne-5\)
\(A=\dfrac{\left(2x+10\right)\left(2x-42\right)}{x+5}\)
Để A = 0, thì:
\(\left[{}\begin{matrix}2x+10=0\\2x-42=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=-10\\2x=42\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\left(ktm\right)\\x=21\left(tm\right)\end{matrix}\right.\)
Vậy ...