Ta có: \(\left(5x-1\right)^2+\left(5x-4\right)\left(5x+4\right)=7\)
\(\Leftrightarrow25x^2-10x+1+25x^2-16-7=0\)
\(\Leftrightarrow25x^2-10x-22=0\)
\(\Leftrightarrow\left(5x\right)^2-2\cdot5x\cdot1+1-23=0\)
\(\Leftrightarrow\left(5x-1\right)^2=23\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-1=\sqrt{23}\\5x-1=-\sqrt{23}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=\sqrt{23}+1\\5x=-\sqrt{23}+1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\sqrt{23}+1}{5}\\x=\frac{-\sqrt{23}+1}{5}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{\sqrt{23}+1}{5};\frac{-\sqrt{23}+1}{5}\right\}\)
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