(Yêu cầu: Xem lại điều kiện x, y < 0 giùm)
Ta có: \(\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{x^2}{9}=\frac{y^2}{16}\)\(\Rightarrow\frac{2x^2}{18}=\frac{y^2}{16}\)
Và 2x2 + y2 = 136
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\frac{2x^2}{18}=\frac{y^2}{16}=\frac{2x^2+y^2}{18+16}=\frac{136}{34}=4\)
Ta có:
\(\frac{2x^2}{18}=4\Rightarrow x^2=\frac{4.18}{2}=36\Rightarrow x=-6\) (vì x < 0)
\(\frac{y^2}{16}=4\Rightarrow y^2=4.16=64\Rightarrow y=-8\) (vì y < 0)
Vậy x = -6; y = -8
\(C1:\)
\(\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{x}{3}.\frac{y}{4}=\left(\frac{x}{3}\right)^2.\left(\frac{y}{4}\right)^2\Rightarrow\frac{xy}{3.4}=\frac{x^2}{3^2}=\frac{y^2}{4^2}\Rightarrow\frac{xy}{12}=\frac{2x^2}{2.9}=\frac{y^2}{16}\Rightarrow\frac{xy}{12}=\frac{2x^2}{18}=\frac{y^2}{16}\)
\(\text{Áp dụng tính chất dãy tỉ số bằng nhau: }\)
\(\Rightarrow\frac{2x^2}{18}=\frac{y^2}{16}=\frac{2x^2+y^2}{18+16}=\frac{136}{54}=4\left(\text{do }2x^2+y^2=136\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{2x^2}{18}=4\\\frac{y^2}{16}=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x^2=4.18=72\\y^2=4.16=64\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2=72:2=36\\\Leftrightarrow\left[{}\begin{matrix}y^2=\left(-8\right)^2\\y^2=8^2\end{matrix}\right.\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x^2=\left(-6\right)^2\\x^2=6^2\end{matrix}\right.\\\left[{}\begin{matrix}y=-8\\y=8\end{matrix}\right.\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=-6\\x=6\end{matrix}\right.\\\left[{}\begin{matrix}y=-8\\y=8\end{matrix}\right.\end{matrix}\right.\)
\(\text{Mà }x,y< 0\Rightarrow x=-6,y=-8\)
\(\text{Vậy }x=-6,y=-8\)