Question

Answer 1

\(ĐK:x\ge0\)

\(Đặt:\sqrt{x}=y^2\ge0\)

\(PT\) \(trở\) \(thành\)

\(\sqrt{y^2+4-4y}+\sqrt{y^2+1-2y}=3\)

\(\Leftrightarrow\left|x-2\right|+\left|y-1\right|=3\)

\(-Với\) \(y< 1\)

\(\Rightarrow2-y+1-y=3\)

\(\Leftrightarrow y=0\) \((T/M)\)

\(\Rightarrow x=0\) \((T/M)\)

\(-Với\) \(1\le x< 2\)

\(\Rightarrow2-y+y-1=3\)

\(\Leftrightarrow1=3\) \((vô lí)\)

\(-Với\) \(y\ge2\)

\(\Rightarrow y-2+y-1=3\)

\(\Leftrightarrow y=3\) \((T/M)\)

\(\Rightarrow y^2=9\)

\(\Leftrightarrow\sqrt{x}=9\)

\(\Leftrightarrow x=81\) \((T/M)\)

Vậy PT có 2 nghiệm phân biệt \(x=0;x=81\)