\(\left(x^2-5x+7\right)^2-\left(2x-5\right)^2=0\)
\(\Leftrightarrow\left(x^2-5x+7-2x+5\right)\left(x^2-5x+7+2x-5\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-3\right)\left(x-2\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\\x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow}\left[{}\begin{matrix}x=1\\x=2\\x=3\\x=4\end{matrix}\right.\)
NPT là :S={1;2;3;4}
\(\Leftrightarrow\left[\left(x^2-5x+7\right)-\left(2x-5\right)\right]\left[\left(x^2-5x+7\right)+\left(2x-5\right)\right]=0\)
\(\Leftrightarrow\left(x^2-7x+12\right)\left(x^2-3x+2\right)=0\)
\(\Leftrightarrow\left(x^2-3x-4x+12\right)\left(x^2-x-2x+2\right)=0\)
\(\Leftrightarrow\left[x\left(x-3\right)-4\left(x-3\right)\right]\left[x\left(x-1\right)-2\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-3\right)\left(x-2\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-3=0\\x-2=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=3\\x=2\\x=1\end{matrix}\right.\left(TM\right)\)
Vậy tập nghiệm của pt đã cho là \(S=\left\{4;3;2;1\right\}\)
