\(\dfrac{x}{3}=\dfrac{y}{4}\Rightarrow\dfrac{x^2}{9}=\dfrac{y^2}{16}\)
Áp dụng tccdtsbn ta có:
\(\dfrac{x^2}{9}=\dfrac{y^2}{16}=\dfrac{x^2+y^2}{9+16}=\dfrac{100}{25}=4\)
\(\Rightarrow\left\{{}\begin{matrix}x^2=4\cdot9=36\\y^2=4\cdot16=64\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\pm6\\y=\pm8\end{matrix}\right.\)
Vậy...........
\(\dfrac{x}{3}=\dfrac{y}{4}\Rightarrow\dfrac{x^2}{9}=\dfrac{y^2}{16}\)
Áp dụng dãy tỉ số = nhau:
\(\dfrac{x^2}{9}=\dfrac{y^2}{16}=\dfrac{x^2+y^2}{25}=4\)
\(\left\{{}\begin{matrix}x^2=36\\y^2=64\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\pm6\\y=\pm8\end{matrix}\right.\)
\(\dfrac{x}{3}=\dfrac{y}{4}\)
\(\Rightarrow\left(\dfrac{x}{3}\right)^2=\left(\dfrac{y}{4}\right)^2\)
\(\Rightarrow\dfrac{x^2}{9}=\dfrac{y^2}{16}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x^2}{9}=\dfrac{y^2}{16}\)
\(=\dfrac{x^2+y^2}{9+16}\)
\(=\dfrac{100}{25}=4\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x^2}{9}=4\Rightarrow x^2=36\Rightarrow x=\pm6\\\dfrac{y^2}{16}=4\Rightarrow y^2=64\Rightarrow x=\pm8\end{matrix}\right.\)
