ĐK: ...
Rút y = 17 - x từ pt dưới thế vào pt trên:
\(\sqrt{x+1}+\sqrt{19-x}=6\) \(\left(ĐK:-1\le x\le19\right)\)
\(\Leftrightarrow\sqrt{x+1}-2+\sqrt{19-x}-4=0\)
\(\Leftrightarrow\left(x-3\right)\left[\frac{1}{\sqrt{x+1}+2}-\frac{1}{\sqrt{19-x}+4}\right]=0\)
\(\Rightarrow x=3\)
đk: \(x\ge-1;y\ge-2\)
đặt: \(\sqrt{x+1}=a;\sqrt{y+2}=b\left(a,b\ge0\right)\)
khi đó: \(a^2+b^2=x+y+3=20\)
ta có hệ:\(\left\{{}\begin{matrix}a+b=6\\a^2+b^2=20\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a=6-b\\\left(6-b\right)^2+b^2=20\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=6-b\\2b^2-12b+16=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}a=6-4=2\\a=6-2=4\end{matrix}\right.\\\left[{}\begin{matrix}b=4\\b=2\end{matrix}\right.\end{matrix}\right.\)(TM)
+) với b = 4; a = 2 => \(\left\{{}\begin{matrix}x=3\\y=14\end{matrix}\right.\)
+) b=2; a=4 => \(\left\{{}\begin{matrix}x=15\\y=2\end{matrix}\right.\)
vậy.......