Question

\(\sqrt{x+2+2\sqrt{x+1}}+\sqrt{x+10-6\sqrt{x+1}}=2\sqrt{x+2-2\sqrt{x+1}}\)

Answer 1

điều kiện xác định : \(x\ge-1\)

ta có : \(\sqrt{x+2+2\sqrt{x+1}}+\sqrt{x+10-6\sqrt{x+1}}=2\sqrt{x+2-2\sqrt{x+1}}\)

\(\Leftrightarrow\sqrt{\left(\sqrt{x+1}+1\right)^2}+\sqrt{\left(\sqrt{x+1}-3\right)^2}=2\sqrt{\left(\sqrt{x+1}-1\right)^2}\) (1)

th1: \(-1\le x< 0\)

\(\Rightarrow\left(1\right)\Leftrightarrow\sqrt{x+1}+1+3-\sqrt{x+1}=2-2\sqrt{x+1}\)

\(\Leftrightarrow2=-2\sqrt{x+1}\) (vô lí)

th2: \(0\le x< 8\)

\(\Rightarrow\left(1\right)\Leftrightarrow\sqrt{x+1}+1+3-\sqrt{x+1}=2\sqrt{x+1}-2\)

\(\Leftrightarrow\sqrt{x+1}=3\Leftrightarrow x=8\) (loại)

th3: \(x\ge8\)

\(\Rightarrow\left(1\right)\Leftrightarrow\sqrt{x+1}+1+\sqrt{x+1}-3=2\sqrt{x+1}-2\)

\(\Leftrightarrow-2=-2\) (luôn đúng)

vậy \(x\ge8\)