Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=k->\left\{{}\begin{matrix}x=2k\\y=3k\end{matrix}\right.\)
Thay vào \(x^2+2y^2=88\)
\(=>\left(2k\right)^2+2.\left(3k\right)^2=88\)
\(=>4.k^2+18.k^2=88\)
\(=>k^2\left(4+18\right)=88\)
\(=>k^2=4\)
\(=>\left[{}\begin{matrix}k=-2\\k=2\end{matrix}\right.\)
Th1: \(k=-2\)
\(=>x=-4;y=-6\)
Th2:\(k=2\)
\(=>x=4;y=6\)
Vậy có 2 cặp \(\left(x;y\right)\) t/m: \(\left(-4;-6\right);\left(4;6\right)\)
Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=k\) (k ∈ N*)
\(\Rightarrow x=2k;y=3k\) (*)
Thay (*) vào biểu thức \(x^2+2y^2=88\) , ta được:
\(\left(2k\right)^2+2\cdot\left(3k\right)^2=88\)
\(\Leftrightarrow4k^2+18k^2=88\)
\(\Leftrightarrow22k^2=88\)
\(\Leftrightarrow k^2=4\Leftrightarrow\left[{}\begin{matrix}k=2\\k=-2\end{matrix}\right.\) (tmđk)
\(+,\) Với \(k=2\) \(\Rightarrow\left\{{}\begin{matrix}x=2\cdot2=4\\y=2\cdot3=6\end{matrix}\right.\) (tm)
\(+,\) Với \(k=-2\)\(\Rightarrow\left\{{}\begin{matrix}x=-2\cdot2=-4\\y=-2\cdot3=-6\end{matrix}\right.\) (tm)
