$\left\{\begin{matrix} (x+2)\sqrt{x+y+4}=x^{3}+x^{2}+y+3 & & \\ (x^{2}+x)\sqrt{x-y+3}=2x^{2}+x+y+1 & & \end{matrix}\right.$
#2
Posted 13-06-2016 - 22:43
Baoriven
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- Điều hành viên OLYMPIC
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- 1426 posts
Thượng úy
Đặt: $a=x^2+x;b=\sqrt{x-y+3},b\geq 0$
Từ phương trình (2) ta được: $(x^2+x)\sqrt{x-y+3}=2(x^2+x)-(x-y+3)+4\Rightarrow ab=2a-b^2+4$
Suy ra: $(b-2)(a+b+2)=0$
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