\(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=\dfrac{x+3}{2}\)
ĐK:\(x\ge 1\)
\(\Leftrightarrow\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}=\dfrac{x+3}{2}\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}=\dfrac{x+3}{2}\)
\(\Leftrightarrow\sqrt{x-1}+1+\left|\sqrt{x-1}-1\right|=\dfrac{x+3}{2}\left(x\ge1\right)\)
\(\Leftrightarrow\sqrt{x-1}+\left|\sqrt{x-1}-1\right|=\dfrac{x+1}{2}\)
*)Xét \(x\le 1\)\(\Rightarrow\sqrt{x-1}-\sqrt{x-1}+1=\dfrac{x+1}{2}\)
\(\Rightarrow\dfrac{x+1}{2}=1\Rightarrow x+1=2\Rightarrow x=1\)
*)Xét \(x>1\)\(\Rightarrow\sqrt{x-1}+\sqrt{x-1}-1=\dfrac{x+1}{2}\)
\(\Rightarrow2\sqrt{x-1}=\dfrac{x+3}{2}\)\(\Rightarrow4\sqrt{x-1}=x+3\)
\(\Rightarrow16\left(x-1\right)=x^2+6x+9\)\(\Rightarrow x=5\)
Vậy \(x=1;x=5\)
