e: Ta có: \(2\sqrt{x^2+7}=2-x\)
\(\Leftrightarrow4x^2+28=x^2-4x+4\)
\(\Leftrightarrow3x^2+4x+24=0\)
\(\text{Δ}=4^2-4\cdot3\cdot24=16-12\cdot24< 0\)
Vì Δ<0 nên phương trình vô nghiệm
a. \(\sqrt{x^2-6x+9}=3x-1\)
<=> \(\sqrt{\left(x-3\right)^2}=3x-1\)
<=> \(|x-3|=3x-1\)
<=> \(\left[{}\begin{matrix}x-3=3x-1\\x-3=-3x+1\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)
b. \(\sqrt{1-4x+4x^2}=5\)
<=> \(\sqrt{\left(1-2x\right)^2}=5\)
<=> \(|1-2x|=5\)
<=> \(\left[{}\begin{matrix}1-2x=5\\1-2x=-5\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)
c. \(\sqrt{36x^2-24x+4}=x\)
<=> \(\sqrt{\left(6x-2\right)^2}=x\)
<=> \(|6x-2|=x\)
<=> \(\left[{}\begin{matrix}6x-2=x\\6x-2=-x\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=0,4\\x=\dfrac{2}{7}\end{matrix}\right.\)
