$\begin{cases}2x^{2}+3=(4x^{2}-2x^{2}y)\sqrt{3-2y}+\frac{4x^{2}+1}{x}\\ \sqrt{2-\sqrt{3-2y}}=\frac{\sqrt[3]{x^{3}+2x^{2}}+x+2}{2x+1}\end{cases}$
*)$(x;y)=(\frac{\sqrt{5}-1}{2};\frac{3+\sqrt{5}}{4})$
thích thì lát up 1 câu, nhìn dễ mak vi diệu @@ –  Confusion 25-06-16 06:57 PM
nhìn thì nguy hiểm chứ k khó j –  phu1279 23-06-16 11:46 AM
Phương trình(1) đưa về xét đơn điệu f(t)=t^3+t cho $\frac{x-1}{x}$và$\sqrt{3-2y}$
thay vô (2) rồi chia cả tử và mẫu của vế phải cho x thì ta lại xét f(a)=a^3+a
cho $\sqrt{1+\frac{1}{x}}$ và $\sqrt[3]{1+\frac{2}{x}}$
Rất tắt nhưng ai k hiểu thì chịu
chủ yếu dùng pp hàm số mà:)) –  Bloody's Rose 23-06-16 04:50 PM

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