ĐKXĐ: \(x\ge-\frac{1}{2}\)
\(x^2-6x+26=6\sqrt{2x+1}\)
\(\Rightarrow2x+1-6\sqrt{2x+1}+x^2-8x+25=0\)
Đặt \(a=\sqrt{2x+1}\left(a\ge0\right)\), ta được pt: a2 - 6a + x2 - 8x + 25 = 0
Có: \(\Delta=36-4\left(x^2-8x+25\right)=-4x^2+32x-64=-4\left(x-4\right)^2\)\(\)
Vì \(\Delta< 0\) nên pt vô nghiệm
Vậy \(x\in\left\{\phi\right\}\)
GT \(\Leftrightarrow x^2-6x+26-6\sqrt{2x+1}=0\)
\(\Leftrightarrow2x+1-2\sqrt{2x+1}\times3+9+x^2-8x+16\)
\(\Leftrightarrow\left(\sqrt{2x+1}-3\right)^2+\left(x-4\right)^2=0\)
suy ra x = 4
