\(\sqrt{1+x}+\sqrt{8-x}=\sqrt{\left(1+x\right).\left(8-x\right)}+3\) ĐK : \(\left\{{}\begin{matrix}x\ge-1\\x\le8\end{matrix}\right.\)
Đặt \(\sqrt{1+x}+\sqrt{8-x}=t\) (ĐK : \(t>0\))
\(\Leftrightarrow t^2=\left(\sqrt{1+x}+\sqrt{8-x}\right)^2\)
\(\Leftrightarrow t^2=1+x+8-x+2.\sqrt{\left(1+x\right).\left(8-x\right)}\)
\(\Leftrightarrow t^2=9+2.\sqrt{\left(1+x\right).\left(8-x\right)}\)
\(\Rightarrow\sqrt{\left(1+x\right).\left(8-x\right)}=\dfrac{t^2-9}{2}\)
Pt trở thành : \(t=\dfrac{t^2-9}{2}+3\Leftrightarrow2t=t^2-9+6\)
\(\Leftrightarrow t^2-2t-3=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=3\left(tm\right)\\t=-1\left(l\right)\end{matrix}\right.\)
Với t = 3 ; ta được :
\(\sqrt{1+x}+\sqrt{8-x}=3\)
\(\Leftrightarrow\left(\sqrt{1+x}+\sqrt{8-x}\right)^2=9\)
\(\Leftrightarrow1+x+8-x+2.\sqrt{\left(1+x\right).\left(8-x\right)}=9\)
\(\Leftrightarrow2.\sqrt{\left(1+x\right).\left(8-x\right)}=0\)
\(\Leftrightarrow\sqrt{\left(1+x\right).\left(8-x\right)}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=8\end{matrix}\right.\)
