ĐKXĐ: \(x^2-3x+1\ge0\)
=>\(x^2-3x+\frac94-\frac54\ge0\)
=>\(\left(x-\frac32\right)^2\ge\frac54\)
=>\(\left[\begin{array}{l}x-\frac32\ge\frac{\sqrt5}{2}\\ x-\frac32\le-\frac{\sqrt5}{2}\end{array}\right.\Rightarrow\left[\begin{array}{l}x\ge\frac{\sqrt5+3}{2}\\ x\le\frac{3-\sqrt5}{2}\end{array}\right.\)
Ta có: \(\sqrt{x^2-3x+1}+\sqrt{x^2-3x+4}=3\)
=>\(\sqrt{x^2-3x+1}-1+\sqrt{x^2-3x+4}-2=0\)
=>\(\frac{x^2-3x+1-1}{\sqrt{x^2-3x+1}+1}+\frac{x^2-3x+4-4}{\sqrt{x^2-3x+4}+2}=0\)
=>\(\left(x^2-3x\right)\left(\frac{1}{\sqrt{x^2-3x+1}+1}+\frac{1}{\sqrt{x^2-3x+2}+2}\right)=0\)
=>\(x^2-3x=0\)
=>x(x-3)=0
=>x=0(nhận) hoặc x=3(nhận)
