hệ phương trình  $\begin{cases}(x^2+x)y^2-4y^2+y+1=0 \\ xy+x^2y^2+1-(4-x^3)y^3=0 \end{cases}$
đơn giản là chia lần lượt y mũ 2 và y mũ 3 là ok –  nguyenlinhkhang1995 28-12-12 08:40 PM
y=0 không là nghiệm nên pt(1) chia cho y^2 chia pt(2) cho y^3đặt $\left\{ \begin{array}{l} x=a\\ \frac{1}{y}=b \end{array} \right.$ $\left\{ \begin{array}{l} a^2 a b^2 b=4\\ a^3 ab^2 a^2b b^3=4 \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} –  nguyenlinhkhang1995 28-12-12 08:39 PM

bị xóa vì bị trùng bởi Đức Vỹ 24-10-13 04:34 PM

Câu hỏi này có nội dung hoàn toàn trùng với câu hỏi đã có.

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