ĐK: \(x\ge3\)
\(pt\Leftrightarrow\sqrt{x-1}=\sqrt{x-2}+\sqrt{x-3}\)
\(\Leftrightarrow x-1=x-2+x-3+2\sqrt{\left(x-2\right)\left(x-3\right)}\)
\(\Leftrightarrow2\sqrt{x^2-5x+6}=4-x\)
\(\Leftrightarrow\left\{{}\begin{matrix}4\left(x^2-5x+6\right)=\left(4-x\right)^2\\x\le4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x^2-12x+8=0\\x\le4\end{matrix}\right.\)
\(\Leftrightarrow x=\dfrac{6+2\sqrt{3}}{3}\left(tm\right)\)
ĐK: \(\begin{cases}x≥1\\x≥2\\x≥3\end{cases} \Leftrightarrow x≥3\)
\(\sqrt{x-1}-\sqrt{x-2}=\sqrt{x-3} \\ \Leftrightarrow \sqrt{x-1}=\sqrt{x-3}+\sqrt{x-2} \\ \Leftrightarrow x-1= x-3+x-2+2\sqrt{(x-2)(x-3)} \\\Leftrightarrow -x+4=2\sqrt{(x-2)(x-3)} \\ \Leftrightarrow x^2-8x+16=4(x-2)(x-3) (ĐK: 3≤x≤4) \\ \Leftrightarrow x^2-8x+16=4x^2-20x+24\\\Leftrightarrow3x^2-12x+8=0 \\ \Leftrightarrow x = \dfrac{6 \pm 2\sqrt3}{3}\)
Vậy \(x=\dfrac{6+2\sqrt3}{3}\).
