\(\sqrt{x+1}-\sqrt{x-2}=\sqrt{x+3}\)
\(\Leftrightarrow\sqrt{x+3}+\sqrt{x+2}=\sqrt{x+1}\)
ĐKXĐ : \(\left\{{}\begin{matrix}x+1\ge0\\x+3\ge0\\x-2\ge0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x\ge-3\\x\ge2\end{matrix}\right.\)\(\Leftrightarrow x\ge2\)
\(pt\Leftrightarrow\left(\sqrt{x+3}+\sqrt{x-2}\right)^2=x+1\)
\(\Leftrightarrow x+3+2.\sqrt{\left(x+3\right).\left(x-2\right)}+x-2=x+1\)
\(\Leftrightarrow2.\sqrt{\left(x+3\right).\left(x-2\right)}=-x\)
\(\Leftrightarrow\sqrt{\left(x+3\right).\left(x-2\right)}=\dfrac{-1}{2}x\)
\(\Leftrightarrow\left(x+3\right).\left(x-2\right)=\dfrac{1}{4}x^2\)
\(\Leftrightarrow4.\left(x+3\right).\left(x-2\right)=x^2\)
\(\Leftrightarrow\left(4x+12\right).\left(x-2\right)-x^2=0\)
\(\Leftrightarrow4x^2-8x+12x-24-x^2=0\)
\(\Leftrightarrow3x^2+4x-24=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-2+\sqrt{76}}{3}\left(\text{thỏa mãn}\right)\\x=\dfrac{-2-\sqrt{76}}{3}\left(\text{loại }\right)\end{matrix}\right.\)
Vậy -..
