điều kiện : \(x\ge1\)
ta có : \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=\dfrac{x+3}{2}\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}\right)^2+2\sqrt{x-1}+1}+\sqrt{\left(\sqrt{x-1}\right)^2-2\sqrt{x-1}+1}=\dfrac{x+3}{2}\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}=\dfrac{x+3}{2}\)
\(\Leftrightarrow\sqrt{x-1}+1+\left|\sqrt{x-1}-1\right|=\dfrac{x+3}{2}\) (1)
th1: \(\sqrt{x-1}-1\ge0\Leftrightarrow x\ge2\)
ta có : \(2\sqrt{x-1}=\dfrac{x+3}{2}\Leftrightarrow4\sqrt{x-1}=x+3\)
\(\Leftrightarrow16\left(x-1\right)=x^2+6x+9\Leftrightarrow x^2-10x+25=0\) \(\Leftrightarrow x=5\left(tmđk\right)\)
th1: \(\sqrt{x-1}-1< 0\Leftrightarrow1\le x< 2\)
ta có : \(2=\dfrac{x+3}{2}\Leftrightarrow4=x+3\Leftrightarrow x=1\left(tmđk\right)\)
vậy \(x=5;x=1\)
