Câu hỏi của ĐTT

Question

Tìm x; y biết $\dfrac{y^2-x^2}{3}=\dfrac{y^2+x^2}{5}$ và $x^{10}.y^{10}=1024$

Nguyễn Việt Lâm · Accepted Answer

$\dfrac{y^2-x^2}{3}=\dfrac{y^2+x^2}{5}=\dfrac{2y^2}{8}=\dfrac{2x^2}{2}$

$\Rightarrow\dfrac{2y^2}{8}=\dfrac{2x^2}{2}\Rightarrow y^2=4x^2$

Lại có $x^{10}.y^{10}=1024\Leftrightarrow x^{10}.\left(y^2\right)^5=1024$

$\Leftrightarrow x^{10}.\left(4x^2\right)^5=1024\Leftrightarrow4^5.x^{10}.x^{10}=1024$

$\Leftrightarrow1024.x^{20}=1024\Rightarrow x^{20}=1\Rightarrow x=\pm1$

$\Rightarrow y^2=4x^2=4\Rightarrow y=\pm2$

Vậy $\left\{{}\begin{matrix}x=\pm1\y=\pm2\end{matrix}\right.$Câu trả lời: $\dfrac{y^2-x^2}{3}=\dfrac{y^2+x^2}{5}=\dfrac{2y^2}{8}=\dfrac{2x^2}{2}$

$\Rightarrow\dfrac{2y^2}{8}=\dfrac{2x^2}{2}\Rightarrow y^2=4x^2$

Lại có $x^{10}.y^{10}=1024\Leftrightarrow x^{10}.\left(y^2\right)^5=1024$

$\Leftrightarrow x^{10}.\left(4x^2\right)^5=1024\Leftrightarrow4^5.x^{10}.x^{10}=1024$

$\Leftrightarrow1024.x^{20}=1024\Rightarrow x^{20}=1\Rightarrow x=\pm1$

$\Rightarrow y^2=4x^2=4\Rightarrow y=\pm2$

Vậy $\left\{{}\begin{matrix}x=\pm1\y=\pm2\end{matrix}\right.$