2<m<4

\(\hept{\begin{cases}\sqrt{12-2x^2}=4+y\\\sqrt{1-2y-y^2}=5-2x\end{cases}}\)(ĐK: \(\hept{\begin{cases}12-2x^2\ge0\\1-2y-y^2\ge0\end{cases}}\))

\(\Rightarrow\hept{\begin{cases}12-2x^2=\left(4+y\right)^2\\1-2y-y^2=\left(5-2x\right)^2\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}y^2+2x^2+8y+4=0\\4x^2+y^2-20x+2y+24=0\end{cases}}\)

\(\Rightarrow2\left(4x^2+y^2-20x+2y+24\right)+\left(y^2+2x^2+8y+4\right)=0\)

\(\Leftrightarrow\left(x+y\right)^2+\left(x-y-4\right)^2+8\left(x-2\right)^2+\left(y+2\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}x=2\\y=-2\end{cases}}\)

Thử lại thỏa mãn. 

Ta có |x + 3| + |7 - x| \(\ge\left|x+3+7-x\right|=\left|10\right|=10\)

Dấu "=" xảy ra <=> \(\left(x+3\right)\left(7-x\right)\ge0\)

Xét các trường hợp

TH1 : \(\hept{\begin{cases}x+3\ge0\\7-x\ge0\end{cases}}\Rightarrow\hept{\begin{cases}x\ge-3\\x\le7\end{cases}}\Rightarrow-3\le x\le7\)(tm)

TH2 \(\hept{\begin{cases}x+3\le0\\7-x\le0\end{cases}}\Rightarrow\hept{\begin{cases}x\le-3\\x\ge7\end{cases}}\left(\text{loại}\right)\)

Vậy \(-3\le x\le7\)là giá trị cần tìm

\(\sqrt{x}+\sqrt{9-x}=\sqrt{-x^2+9x+9}\)(ĐK: \(0\le x\le9\))

\(\Leftrightarrow x+9-x+2\sqrt{x\left(9-x\right)}=-x^2+9x+9\)

\(\Leftrightarrow4x\left(9-x\right)=x^2\left(9-x\right)^2\)

\(\Leftrightarrow\orbr{\begin{cases}x\left(9-x\right)=0\\x\left(9-x\right)=4\end{cases}}\)

- \(x\left(9-x\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=9\end{cases}}\).

- \(x\left(9-x\right)=4\Leftrightarrow\orbr{\begin{cases}x=\frac{9+\sqrt{65}}{2}\\x=\frac{9-\sqrt{65}}{2}\end{cases}}\)