\[\left\{\begin{matrix} (x+\sqrt{x^{2}+4})(y+\sqrt{y^{2}+1})=2\\ 12y^{2}-10y+2=2 \sqrt[3]{x^{3}+1} \end{matrix}\right.\]
2)
\[\left\{\begin{matrix} y+\sqrt{y^{2}-2y+5}=3x+\sqrt{x^{2}+4}\\y^{2} -x^{2}-3y+3x+1=0 \end{matrix}\right.\]
3)
\[\left\{\begin{matrix} x^{2}+y^{2}-y=(2x+1)(y-1)\\ \sqrt{3x-8}-\sqrt{y}=\frac{5}{x+y+2} \end{matrix}\right.\]
4)
\[\left\{\begin{matrix} x+3y+1=y^{2}-\frac{1}{y}+\frac{3x+y}{\sqrt{x+1}}\\ \sqrt{9y-2} + \sqrt[3]{7x+2y+2}=2y+3 \end{matrix}\right.\]
6)
\[\left\{\begin{matrix} 2x^{2}+x+\sqrt{x+2}=2y^2+y+\sqrt{2y+1}\\ x^{2}+2y^2-2x+y-2=0 \end{matrix}\right.\]
7)
\[\left\{\begin{matrix} 3(x+y)-2\sqrt{xy}=8\\\sqrt{x+7}+ \sqrt{y+7}=6 \end{matrix}\right.\]
8)
\[\left\{\begin{matrix} \sqrt{x} + \sqrt{y} = 4\\ \sqrt{x+7} + \sqrt{y+7} = 6 \end{matrix}\right.\]
Bài viết đã được chỉnh sửa nội dung bởi tthandb: 19-07-2014 - 21:35