1,$\left\{\begin{matrix}x^{4}-y^{4}=\frac{3}{4y}-\frac{1}{2x} & \\ (x^{2}-y^{2})^{5}+5=0\end{matrix}\right.$
2,$\left\{\begin{matrix} (\sqrt{y}+1)^{2}+\frac{y^{2}}{x}=y^{2}+2.\sqrt{x-2} & \\x+\frac{x-1}{y}+\frac{y}{x}=y^{2}+y & \end{matrix}\right.$
3,$8x^{3}-12x+7x=(x+1).\sqrt[3]{3x^{2}-2}$
4,$x^{3}-5x^{2}+4x-5=(1-2x).\sqrt[3]{6x^{2}-2x+7}$
5,$\frac{1}{\sqrt{x+3}}+\frac{1}{\sqrt{3x+1}}=\frac{2}{1+\sqrt{x}}$
6,$\frac{1}{x+\sqrt{x^{2}-1}}=\frac{1}{4x}+\frac{3x}{2x^{2}+2}$
7,$\left\{\begin{matrix}\sqrt{2x-3}-\sqrt{y}=2x-6 & \\ x^{3}+y^{3}+7.(x+y)xy=8xy\sqrt{2.(x^{2}+y^{2})} & \end{matrix}\right.$
8,$\left\{\begin{matrix}4x^{2}=(\sqrt{x^{2}+1}+1)(x^{2}-y^{2}+3y-2) &\\(x^{2}+y^{2})^{2}+1=x^{2}+2y& \end{matrix}\right.$
Bài viết đã được chỉnh sửa nội dung bởi ecchi123: 18-10-2015 - 23:20