\(\sqrt{10-x}+\sqrt{x+3}=5\left(1\right)\)
Bình phương 2 vế, ta được:
\(\left(1\right)\Rightarrow10-x+x+3+2\sqrt{\left(10-x\right)\left(x+3\right)}=5^2\)
\(\Leftrightarrow2\sqrt{10x+30-x^2-3x}=25-13\)
\(\Leftrightarrow2\sqrt{10x+30-x^2-3x}=12\left(2\right)\)
Tiếp tục bình phương 2 vế, ta có:
\(\left(2\right)\Rightarrow4\left(10x+30-x^2-3x\right)=12^2\)
\(\Leftrightarrow40x+120-4x^2-12x=144\)
\(\Leftrightarrow28x-4x^2-24=0\)
\(\Leftrightarrow4x-4x^2+24x-24=0\)
\(\Leftrightarrow4x\left(1-x\right)-24\left(1-x\right)=0\)
\(\Leftrightarrow4\left(x-6\right)\left(1-x\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}x-6=0\\1-x=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=6\\x=1\end{cases}}}\)
Vậy \(S=\left\{1;6\right\}\)
(Chúc bạn học tốt nhá!)
