Ta có: \(\frac{x^2+y^2}{10}=\frac{x^2-2y^2}{7}\)
\(\Leftrightarrow7x^2+7y^2=10x^2-20y^2\)
\(\Leftrightarrow27y^2=3x^2\)
\(\Leftrightarrow9y^2=x^2\)
\(\Leftrightarrow81y^4=x^4\)
Ta lại có: \(x^4y^4=81\)
\(\Rightarrow81y^4.y^4=81\)
\(\Leftrightarrow y^8=1\)
\(\Leftrightarrow\left[{}\begin{matrix}y=1\\y=-1\end{matrix}\right.\)
\(y=\pm1\Rightarrow x^2=9y^2=9\)
\(\Rightarrow x=\pm3\)
Pt có nghiệm \(\left(x,y\right)=\left\{\left(3;1\right);\left(-3;1\right);\left(3;-1\right);\left(-3;-1\right)\right\}\)
Ta có: \(\frac{x^2+y^2}{10}=\frac{x^2-2y^2}{7}.\)
Đặt \(\left\{{}\begin{matrix}x^2=a\left(a\ge0\right)\\y^2=b\left(b\ge0\right)\end{matrix}\right.\)
\(\Rightarrow\frac{a+b}{10}=\frac{a-2b}{7}\) và \(a^2.b^2=81.\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được:
\(\frac{a+b}{10}=\frac{a-2b}{7}=\frac{a+b-\left(a-2b\right)}{10-7}=\frac{a+b-a+2b}{3}=\frac{3b}{3}=b\) (1).
\(\frac{a+b}{10}=\frac{2a+2b}{20}=\frac{a-2b}{7}=\frac{2a+2b+a-2b}{20+7}=\frac{3a}{27}=\frac{a}{9}\) (2).
Từ (1) và (2) \(\Rightarrow\frac{a}{9}=b.\)
\(\Rightarrow a=9b.\)
Vì \(a^2.b^2=81\)
\(\Rightarrow\left(9b\right)^2.b^2=81\)
\(\Rightarrow81b^2.b^2=81\)
\(\Rightarrow81.b^4=81\)
\(\Rightarrow b^4=81:81\)
\(\Rightarrow b^4=1\)
\(\Rightarrow b=1\) (vì \(b\ge0\)).
Mà \(a=9b\)
\(\Rightarrow a=9.1\)
\(\Rightarrow a=9.\)
\(\Rightarrow\left\{{}\begin{matrix}x^2=9\\y^2=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\\\left[{}\begin{matrix}y=1\\y=-1\end{matrix}\right.\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(3;1\right),\left(-3;-1\right).\)
Chúc bạn học tốt!
