\(\dfrac{x}{5}=\dfrac{y}{4}\)
\(\Leftrightarrow\dfrac{x^2}{25}=\dfrac{y^2}{16}\)
Theo t,c dãy tỉ số bằng nhau ta có :
\(\dfrac{x^2}{25}=\dfrac{y^2}{16}=\dfrac{x^2-y^2}{25-16}=\dfrac{36}{9}=4\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x^2}{25}=4\\\dfrac{y^2}{16}=4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x^2=100\\y^2=64\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=10\\x=-10\end{matrix}\right.\\\left[{}\begin{matrix}y=8\\y=-8\end{matrix}\right.\end{matrix}\right.\)
Vậy ..
\(x^2-y^2=36\Leftrightarrow\left(x+y\right)\left(x-y\right)=36\)(1)
\(\dfrac{x}{5}=\dfrac{y}{4}=\dfrac{x+y}{9}=\dfrac{x-y}{1}\)(2)
Ta có: \(\left\{{}\begin{matrix}\left(x+y\right)\left(x-y\right)=36\\\dfrac{x+y}{9}=\dfrac{x-y}{1}\end{matrix}\right.\)
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