ĐKXĐ: \(x\ge0\)
Đặt \(3+\sqrt{x}=t>0\Rightarrow3=t-\sqrt{x}\)
Phương trình trở thành:
\(x+\sqrt{t}=t-\sqrt{x}\)
\(\Leftrightarrow x-t+\sqrt{x}+\sqrt{t}=0\)
\(\Leftrightarrow\left(\sqrt{x}+\sqrt{t}\right)\left(\sqrt{x}-\sqrt{t}\right)+\sqrt{x}+\sqrt{t}=0\)
\(\Leftrightarrow\left(\sqrt{x}+\sqrt{t}\right)\left(\sqrt{x}-\sqrt{t}+1\right)=0\)
\(\Leftrightarrow\sqrt{x}+1=\sqrt{t}\)
\(\Leftrightarrow\sqrt{x}+1=\sqrt{3+\sqrt{x}}\)
\(\Leftrightarrow x+2\sqrt{x}+1=3+\sqrt{x}\)
\(\Leftrightarrow x+\sqrt{x}-2=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)=0\)
\(\Leftrightarrow x=1\)
Đúng 2
Bình luận (0)
