\(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{3}\Rightarrow\dfrac{x^2}{25}=\dfrac{y^2}{49}=\dfrac{z^2}{9}\)
a/dụng t/c của dãy tỉ số = nhau có:
\(\dfrac{x^2}{25}=\dfrac{y^2}{49}=\dfrac{z^2}{9}=\dfrac{x^2+y^2-z^2}{25+49-9}=\dfrac{585}{65}=9\)
\(\Rightarrow\left\{{}\begin{matrix}x^2=9\cdot25=225\\y^2=9\cdot49=441\\z^2=9\cdot9=81\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=15\\x=-15\end{matrix}\right.\\\left[{}\begin{matrix}y=21\\y=-21\end{matrix}\right.\\\left[{}\begin{matrix}z=9\\z=-9\end{matrix}\right.\end{matrix}\right.\)
Vậy \(\left(x;y;z\right)=\left(15;21;9\right);\left(-15;-21;-9\right)\)
Ta có: \(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{3}\)
\(\Rightarrow\left(\dfrac{x}{5}\right)^2=\left(\dfrac{y}{7}\right)^2=\left(\dfrac{z}{3}\right)^2\)
\(\Rightarrow\dfrac{x^2}{25}=\dfrac{y^2}{49}=\dfrac{z^2}{9}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{x^2}{25}=\dfrac{y^2}{49}=\dfrac{z^2}{9}=\dfrac{x^2+y^2-z^2}{25+49-9}=\dfrac{585}{65}=9\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x^2}{25}=9\Rightarrow\dfrac{x}{5}=3\Rightarrow x=3.5=15\\\dfrac{y^2}{36}=9\Rightarrow\dfrac{y}{6}=3\Rightarrow y=3.6=18\\\dfrac{z^2}{9}=9\Rightarrow\dfrac{z}{3}=3\Rightarrow z=3.3=9\end{matrix}\right.\)
Vậy x = 15; y = 18; z = 9.
