\(\hept{\begin{cases}m=\sqrt{x+1}+\sqrt{6-y}\left(1\right)\\m=\sqrt{6-x}-\sqrt{1+y}\left(2\right)\end{cases}}\)

ĐKXĐ : \(-1\le x\le6\); \(-1\le y\le6\)

( 1 ) - ( 2 ) , ta được :

\(\sqrt{x+1}-\sqrt{6-x}+\sqrt{6-y}+\sqrt{1+y}=0\)

\(\Leftrightarrow\sqrt{6-x}-\sqrt{x+1}=\sqrt{6-y}+\sqrt{1+y}\)

\(\Leftrightarrow\hept{\begin{cases}\sqrt{6-x}\ge\sqrt{x+1}\\6-x-2\sqrt{\left(6-x\right)\left(x+1\right)}+x+1=6-y+2\sqrt{\left(6-y\right)\left(y+1\right)}+y+1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x\le\frac{5}{2}\\-\sqrt{\left(6-x\right)\left(x+1\right)}=\sqrt{\left(6-y\right)\left(y+1\right)}\end{cases}}\)

Ta thấy VP \(\le\)0 ; VT \(\ge\)0 nên VT = VP = 0

\(\Rightarrow\orbr{\begin{cases}x=-1;y=-1\\x=-1;y=6\end{cases}}\)

với x = y = -1 thì m = \(\sqrt{7}\)

với x = -1 ; y = 6 thì m = 0

Vậy m = \(\sqrt{7}\)hoặc m = 0 thì hệ có nghiệm duy nhất

\(\hept{\begin{cases}\sqrt{x+1}+\sqrt{9-y}=m\\\sqrt{y+1}+\sqrt{9-x}=m\end{cases}}\)

<=>\(\hept{\begin{cases}\sqrt{x+1}+\sqrt{9-y}=m\\\sqrt{y+1}+\sqrt{9-x}=\sqrt{x+1}+\sqrt{9-y}\left(=m\right)\end{cases}}\)

<=>\(\hept{\begin{cases}\sqrt{x+1}+\sqrt{9-y}=m\\y+1+9-x+2\sqrt{\left(y+1\right).\left(9-x\right)}=x+1+9-y+2\sqrt{\left(x+1\right).\left(9-y\right)}\end{cases}}\)

<=>\(\hept{\begin{cases}\sqrt{x+1}+\sqrt{9-y}=m\\2\sqrt{\left(y+1\right).\left(9-x\right)}=2\sqrt{\left(x+1\right).\left(9-y\right)}\end{cases}}\)

<=>\(\hept{\begin{cases}\sqrt{x+1}+\sqrt{9-y}=m\\\sqrt{9y-x+9-xy}=\sqrt{9x-y+9-xy}\end{cases}}\)

<=>\(\hept{\begin{cases}\sqrt{x+1}+\sqrt{9-y}=m\\9y-x+9-xy=9x-y+9-xy\end{cases}}\)

<=>\(\hept{\begin{cases}\sqrt{x+1}+\sqrt{9-y}=m\\10x-10y=0\end{cases}}\)

<=>\(\hept{\begin{cases}\sqrt{x+1}+\sqrt{9-y}=m\\x=y\end{cases}}\)

<=>\(\hept{\begin{cases}\sqrt{x+1}+\sqrt{9-x}=m\\x=y\end{cases}}\)

<=>\(\hept{\begin{cases}x+1+9-x+2\sqrt{\left(x+1\right).\left(9-x\right)}=m^2\\x=y\end{cases}}\)

<=>\(\hept{\begin{cases}2\sqrt{\left(x+1\right).\left(9-x\right)}=m^2-10\\x=y\end{cases}}\)

<=>\(\hept{\begin{cases}4\left(-x^2+8x+9\right)=m^4-20m^2+100\\x=y\end{cases}}\)

<=>\(\hept{\begin{cases}-4x^2+32x+36=m^4-20m^2+100\\x=y\end{cases}}\)

<=>\(\hept{\begin{cases}-4x^2+32x-m^4+20m^2-64=0\left(1\right)\\x=y\end{cases}}\)

Xét (1)

-4x²+32x-m⁴+20m²-64=0

tính delta rồi xét trường hợp nghiệm duy nhất là ra