Ta có : \(\sqrt{1-10x+25x^2}=4x\) ( Điều kiện : \(x\ge0\) )
\(\Rightarrow\sqrt{\left(5x-1\right)^2}=4x\)
\(\Rightarrow\left|5x-1\right|=4x\)
\(\Rightarrow\left[\begin{array}{nghiempt}5x-1=4x\\5x-1=-4x\end{array}\right.\Rightarrow\left[\begin{array}{nghiempt}5x-4x=1\\5x+4x=1\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=1\\9x=1\end{array}\right.\Rightarrow\left[\begin{array}{nghiempt}x=1\\x=\frac{1}{9}\end{array}\right.\)
Vậy \(x\in\left\{1;\frac{1}{9}\right\}\).
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