ĐKXĐ: \(x\ge1\)
\(\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1+2\sqrt{x-1}+1}=\dfrac{x+3}{2}\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}=\dfrac{x+3}{2}\)
\(\Leftrightarrow\sqrt{x-1}+1+\left|\sqrt{x-1}-1\right|=\dfrac{x+3}{2}\) (1)
TH1: \(1\le x\le2\Rightarrow\sqrt{x-1}\le1\)
\(\left(1\right)\Leftrightarrow\sqrt{x-1}+1+1-\sqrt{x-1}=\dfrac{x+3}{2}\)
\(\Leftrightarrow\dfrac{x+3}{2}=2\Leftrightarrow x+3=4\Leftrightarrow x=1\)
TH2: \(x>2\Rightarrow\sqrt{x-1}>1\)
\(\left(1\right)\Leftrightarrow\sqrt{x-1}+1+\sqrt{x-1}-1=\dfrac{x+3}{2}\)
\(\Leftrightarrow2\sqrt{x-1}=x+3\Leftrightarrow4\left(x-1\right)=\left(x+3\right)^2\)
\(\Leftrightarrow x^2+2x+13=0\) (vô nghiệm)
Vậy pt có nghiệm duy nhất \(x=1\)
