đkxđ: -3≤x≤2
\(\sqrt{x+3}-\sqrt{2-x}=1\)
\(\Leftrightarrow\sqrt{x+3}=1+\sqrt{2-x}\)
\(\Leftrightarrow x+3=1+2\sqrt{2-x}+2-x\)(bình phương 2 vế)
\(\Leftrightarrow2x-2\sqrt{2-x}=0\)
\(\Leftrightarrow2\left(x-\sqrt{2-x}\right)=0\)
\(\Leftrightarrow x-\sqrt{2-x}=0\)
\(\Leftrightarrow x=\sqrt{2-x}\)\(\Leftrightarrow x^2=2-x\)
\(\Leftrightarrow x^2+x-2=0\)
\(\Leftrightarrow x^2-x+2x-2=0\)
\(\Leftrightarrow x\left(x-1\right)+2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
Thử lại thấy x = -2 không thỏa mãn
Vậy pt có nghiệm x = 1
