\left\{\begin{matrix}
4xy +4(x^{2}+y^{2})+\frac{3}{(x+y)^{2}} & & \\
2x+\frac{1}{x+y}=3 & &
\end{matrix}\right.
\left\{\begin{matrix} x^{4}-4x^{2}+y^{2}-6y+9=0 & & \\ x^{2}y+x^{2}+2y=-22
Started By skykute, 11-02-2016 - 12:40
#1
Posted 11-02-2016 - 12:40
#2
Posted 11-02-2016 - 12:43
#3
Posted 11-02-2016 - 14:57
Đặt $x^2=a$, thay vào ta có:
$\begin{cases} & a^2-4a+y^2-6y+9=0 \\ & ay+a+2y+22=0 \end{cases}$
$PT(1)+PT(2) \iff a^2+y^2+2ay-2a-2y+53=0$
$\iff (a+y)^2-2(a+y)+53=0$ (PT vô nghiệm)
Vậy hệ vô nghiệm.
Edited by leminhnghiatt, 11-02-2016 - 14:58.
Don't care
1 user(s) are reading this topic
0 members, 1 guests, 0 anonymous users