\(\left(x^2-25\right)^2-\left(x-5\right)^2=0\)
\(\Leftrightarrow\left[\left(x-5\right)\left(x+5\right)\right]^2-\left(x-5\right)^2=0\)
\(\Leftrightarrow\left(x-5\right)^2\cdot\left(x+5\right)^2-\left(x-5\right)^2=0\)
\(\Leftrightarrow\left(x-5\right)^2\cdot\left[\left(x+5\right)^2-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-5\right)^2=0\\\left(x+5\right)^2-1=0\Rightarrow\left(x+5\right)^2=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x+5=1\\x+5=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-4\\x=-6\end{matrix}\right.\)
Vậy: x= 5; x=-4 hoặc x=-6
\(\left(x^2-25\right)^2-\left(x-5\right)^2=0\)
\(\Leftrightarrow\left(x-5\right)^2\left(x+5\right)^2-\left(x-5\right)^2=0\)
\(\Leftrightarrow\left(x-5\right)^2\left[\left(x+5\right)^2-1\right]=0\)
\(\Leftrightarrow\left(x-5\right)^2\left(x+5-1\right)\left(x+5+1\right)=0\)
\(\Leftrightarrow\left(x-5\right)^2\left(x+4\right)\left(x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-5\right)^2=0\\x+4=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-4\\x=-6\end{matrix}\right.\)
Vậy \(x=-6;x=-4;x=5\)
