Câu 1:
ĐK: \(x\in\mathbb{R}\)
\(\sqrt{x^2-2x+5}=x^2-2x-1=x^2-2x+5-6\)
Đặt \(\sqrt{x^2-2x+5}=t(t\geq 0)\). PT trở thành:
\(t=t^2-6\)
\(\Leftrightarrow t^2-t-6=0\Leftrightarrow (t-3)(t+2)=0\)
\(\Leftrightarrow \left[\begin{matrix} t=3\\ t=-2\end{matrix}\right.\). Vì $t\geq 0$ nên $t=3$
Do đó: \(\sqrt{x^2-2x+5}=3\Rightarrow x^2-2x+5=9\)
\(\Rightarrow x^2-2x-4=0\Rightarrow x=1\pm \sqrt{5}\)
Vậy........
Câu 2:
ĐK: \(x\in\mathbb{R}\)
Ta có: \(x^2-4x-6=\sqrt{2x^2-8x+12}\)
\(\Rightarrow 2x^2-8x-12=2\sqrt{2x^2-8x+12}\)
\(\Leftrightarrow (2x^2-8x+12)-24-2\sqrt{2x^2-8x+12}=0\)
Đặt \(\sqrt{2x^2-8x+12}=t(t\geq 0)\). PT trở thành:
\(t^2-24-2t=0\)
\(\Leftrightarrow (t-6)(t+4)=0\Rightarrow \left[\begin{matrix} t=6\\ t=-4\end{matrix}\right.\)
Mà \(t\geq 0\Rightarrow t=6\)
Do đó: \(\sqrt{2x^2-8x+12}=6\Rightarrow 2x^2-8x+12=36\)
\(\Rightarrow x^2-4x-12=0\Rightarrow \left[\begin{matrix} x=6\\ x=-2\end{matrix}\right.\)
Vậy...........
Câu 3:
ĐK: \(x\geq -2\)
\(\sqrt{2-\sqrt{2+x}-x}=0\Rightarrow 2-\sqrt{2+x}-x=0\)
\(\Leftrightarrow 4-\sqrt{2+x}-(x+2)=0\)
Đặt \(\sqrt{2+x}=t(t\geq 0)\)
PT trở thành:
\(4-t-t^2=0\)
\(\Leftrightarrow t^2+t-4=0\Rightarrow t=\frac{-1\pm \sqrt{17}}{2}\)
Vì \(t\ge 0\Rightarrow t=\frac{-1+\sqrt{17}}{2}\)
\(\Rightarrow x=t^2-2=\frac{5-\sqrt{17}}{2}\)
Câu 4:
ĐK: \(x\geq -5\)
\(\sqrt{5-\sqrt{5+x}}=0\)
\(\Rightarrow 5-\sqrt{5+x}=0\)
\(\Rightarrow \sqrt{5+x}=5\Rightarrow 5+x=25\Rightarrow x=20\)
(thỏa mãn)
