\(\left(\sqrt{x+5}-\sqrt{x-2}\right)\left(1+\sqrt{x^2+3x-10}\right)=7\)(đkxđ: \(x\ge2\))
\(\Leftrightarrow\left(\sqrt{x+5}-\sqrt{x-2}\right)\left(1+\sqrt{\left(x+5\right)\left(x-2\right)}\right)=7\)(1)
đặt \(\left\{{}\begin{matrix}\sqrt{x+5}=a\left(đk:a\ge0\right)\\\sqrt{x-2}=b\left(đk:b\ge0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a^2=x+5\\b^2=x-2\end{matrix}\right.\)\(\Leftrightarrow a^2-b^2=7\)(2)
khi đó: \(\left(1\right)\Leftrightarrow\left(a-b\right)\left(1+ab\right)=7\)(3)
từ (2) và (3) ta có : \(\left(a-b\right)\left(1+ab\right)=a^2-b^2\)
\(\Leftrightarrow1+ab=\frac{\left(a-b\right)\left(a+b\right)}{a-b}=a+b\)
\(\Leftrightarrow ab+1-a-b=0\)
\(\Leftrightarrow\left(a-1\right)\left(b-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=1\\b=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+5}=1\\\sqrt{x-2}=1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-4\left(loại\right)\\x=3\left(tm\right)\end{matrix}\right.\)
