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: a2 - 6a + x2 - 8x + 25 = 0
Ta có: \(\Delta'=\left(-3\right)^2-x^2+8x-25=-x^2+8x-16=\left(4-x\right)^2\Rightarrow\sqrt{\Delta'}=4-x\)\(\Rightarrow\left[\begin{array}{nghiempt}a=7-x\\a=x-1\end{array}\right.\)
+) Với a = 7 - x => \(\sqrt{2x+1}=7-x\Rightarrow2x+1=49-14x+x^2\Rightarrow x^2-16x+48=0\)=> x = 4 , x = 12
+) Với a = x - 1 => \(\sqrt{2x+1}=x-1\Rightarrow2x+1=x^2-2x+1\Rightarrow x^2-4x=0\Rightarrow\left[\begin{array}{nghiempt}x=0\\x=4\end{array}\right.\)
Vậy x = 0, x = 4, x = 12
