\(\left\{{}\begin{matrix}\dfrac{x^2.y^2}{10}=\dfrac{x^2-2y^2}{7}\\x^4.y^4=81\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}7.x^2+7.y^2=10.x^2-20.y^2\\\left(x^2.y^2\right)^2=81\end{matrix}\right.\leftrightarrow\left\{{}\begin{matrix}3.x^2=27.y^2\\x^2.y^2=9\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2=9.y^2\\x^2.y^2=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2=9.y^2\\9.y^2.y^2=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2=9.y^2\\\left[{}\begin{matrix}y=1\\y=-1\end{matrix}\right.\end{matrix}\right.\)
(+) \(\left[{}\begin{matrix}\left\{{}\begin{matrix}y=1\\x^2=9\end{matrix}\right.\\\left\{{}\begin{matrix}y=-1\\x^2=9\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}y=1\\x=-9\end{matrix}\right.\\\left\{{}\begin{matrix}y=1\\x=9\end{matrix}\right.\\\left\{{}\begin{matrix}y=-1\\x=-9\end{matrix}\right.\\\left\{{}\begin{matrix}y=-1\\x=9\end{matrix}\right.\end{matrix}\right.\)
Vậy y=1 , x=-9 y=1 , x=9
y=-1 , x=-9 y=-1 , x=9
Đặt \(x^2=a\)(a≥0),\(y^2=b\)(b≥0)
Ta có:\(\dfrac{a+b}{10}=\dfrac{a-2b}{7}vàa^2b^2=81\)
\(\dfrac{a+b}{10}=\dfrac{a-2b}{7}=\dfrac{\left(a+b\right)-\left(a-2b\right)}{10-7}=\dfrac{3b}{3}=b\)(1)
\(\dfrac{a+b}{10}=\dfrac{a-2b}{7}=\dfrac{2a+2b}{20}=\dfrac{\left(2a+2b\right)+\left(a-2b\right)}{20+7}=\dfrac{3a}{27}=\dfrac{a}{9}\left(2\right)\)
Từ (1) và (2) ⇒\(\dfrac{a}{9}=b\)⇒a=9b
Do \(a^2b^2=81nên\left(9b\right)^2b^2=81\)⇒\(b^4=1\)⇒b=2(Vì b≥0)
Suy ra :a=9.1=9 mà x2=a;y2=b⇒ x2=9 và y2=1
⇒xϵ{3;-3} và yϵ{1;-1}
Đặt \(x^2=a\)(a lớn hơn hoặc bằng 0)
\(y^2=b\)(b lớn hơn hoặc bằng 0)
Ta có: \(\dfrac{a+b}{10}=\dfrac{a-2b}{7}\) và \(a^2b^2=81\)
\(\dfrac{a+b}{10}=\dfrac{a-2b}{7}=\dfrac{\left(a+b\right)-\left(a-2b\right)}{10-7}\)
=\(\dfrac{3b}{3}=b\)(1)
\(\dfrac{a+b}{10}=\dfrac{a-2b}{7}=\dfrac{2a+2b}{20+7}=\dfrac{\left(2a+2b\right)+\left(a-2b\right)}{20+7}\)
=\(\dfrac{3a}{27}=\dfrac{a}{9}\)(2)
Từ (1) và (2)=> \(\dfrac{a}{9}=b\)=>a=9b
