Question

Tìm x , y và z biết :

\(\begin{cases} \dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}\\ 2x^2+y^2-z^2=9 \end{cases}\)

Answer 1

Ta có: \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}\) => \(\left(\dfrac{x}{3}\right)^2=\left(\dfrac{y}{4}\right)^2=\left(\dfrac{z}{5}\right)^2\)

=> \(\dfrac{x^2}{9}=\dfrac{y^2}{16}=\dfrac{z^2}{25}\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\dfrac{x^2}{9}=\dfrac{y^2}{16}=\dfrac{z^2}{25}=\dfrac{2x^2+y^2-z^2}{2.9+16-25}=\dfrac{9}{18+16-25}=\dfrac{9}{9}=1\)

=> \(\left\{{}\begin{matrix}\dfrac{x^2}{9}=1\Rightarrow\dfrac{x}{3}=1\Rightarrow x=3\\\dfrac{y^2}{16}=1\Rightarrow\dfrac{y}{4}=1\Rightarrow y=4\\\dfrac{z^2}{25}=1\Rightarrow\dfrac{z}{5}=1\Rightarrow z=5\end{matrix}\right.\)

Vậy x = 3, y = 4, z = 5