\(ĐKXĐ:-3\le x\le6\)
\(\sqrt{x+3}+\sqrt{6-x}=3+\sqrt{\left(x+3\right)\left(6-x\right)}\)
Đặt \(\sqrt{x+3}+\sqrt{6-x}=t\left(t>0\right)\)
\(\Rightarrow t^2=x+3+6-x+2\sqrt{\left(x+3\right)\left(x-6\right)}\)
\(\Rightarrow t^2=9+2\sqrt{\left(x+3\right)\left(x-6\right)}\)
\(\Rightarrow\sqrt{\left(x+3\right)\left(x-6\right)}=\dfrac{t^2-9}{2}\) (1)
Từ (1); pttt :\(t=3+\dfrac{t^2-9}{2}\)
\(\Leftrightarrow2t=6+t^2-9\)
\(\Leftrightarrow t^2-2t-3=0\)
\(\Leftrightarrow\left(t-3\right)\left(t+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=3\left(tm\right)\\t=-1\left(ktm\right)\end{matrix}\right.\)
Thay t = 3 vào (1) , ta được :
\(\sqrt{\left(x+3\right)\left(6-x\right)}=\dfrac{3^2-9}{2}\)
\(\Leftrightarrow\sqrt{\left(x+3\right)\left(6-x\right)}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\6-x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\left(tm\right)\\x=6\left(tm\right)\end{matrix}\right.\)
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