Question

Cho x,y ∈Q, x,y khác 0 thỏa mãn x³+y³=2x²y²

^{Chứng minh rầng :A=\(\sqrt{1-\dfrac{1}{xy}}\)} là số hữu tỉ

Answer 1

Lời giải:

Ta có:

\(x^3+y^3=2x^2y^2\)

\(\Rightarrow \frac{1}{x^3}+\frac{1}{y^3}=\frac{2}{xy}\) (chia 2 vế cho $x^3y^3$)

\(\Rightarrow (\frac{1}{x^3}+\frac{1}{y^3})^2=\frac{4}{x^2y^2}\)

\(\Leftrightarrow \frac{1}{x^6}+\frac{1}{y^6}+\frac{2}{x^3y^3}=\frac{4}{x^2y^2}\)

\(\Leftrightarrow \frac{1}{x^6}+\frac{1}{y^6}-\frac{2}{x^3y^3}=\frac{4}{x^2y^2}-\frac{4}{x^3y^3}\)

\(\Leftrightarrow (\frac{1}{x^3}-\frac{1}{y^3})^2=\frac{4(xy-1)}{x^3y^3}\)

\(\Rightarrow \frac{xy-1}{xy}=\frac{1}{4}x^2y^2(\frac{1}{x^3}-\frac{1}{y^3})^2\)

\(\Rightarrow A=\frac{1}{2}|xy(\frac{1}{x^3}-\frac{1}{y^3})|\in\mathbb{Q}\) do \(x,y\in\mathbb{Q}\)

Ta có đpcm.