Question

\(x:y:z=3:7:5\)và \(x^2-3y^2+z^2=198\)

Answer 1

\(x:y:z=3:7:5\Rightarrow\dfrac{x}{3}=\dfrac{y}{7}=\dfrac{z}{5}\)

Đặt \(\dfrac{x}{3}=\dfrac{y}{7}=\dfrac{z}{5}=k\Rightarrow x=3k;y=7k;z=5k\)

\(x^2-3y^2+z^2=198\\ \Rightarrow\left(3k\right)^2-3\left(7k\right)^2+\left(5k\right)^2=198\\ \Rightarrow9k^2-147k^2+25k^2=198\\ \Rightarrow-113k^2=198\\ \Rightarrow k^2=\dfrac{-198}{113}\left(vô.lí\right)\)

Vậy ko có x,y,z thỏa mãn đề bài

Answer 2

\(x:y:z=3:7:5\Rightarrow\dfrac{x}{3}=\dfrac{y}{7}=\dfrac{z}{5}\Rightarrow\dfrac{x^2}{9}=\dfrac{y^2}{49}=\dfrac{z^2}{25}\)

Áp dụng t/c dtsbn ta có:

\(\dfrac{x^2}{9}=\dfrac{y^2}{49}=\dfrac{z^2}{25}=\dfrac{x^2-3y^2+z^2}{9-3.49+25}=\dfrac{198}{-113}\)

\(\dfrac{x^2}{9}=\dfrac{198}{-113}\Rightarrow x^2=\dfrac{-1782}{113}\left(vô.lí\right)\\ \dfrac{y^2}{49}=\dfrac{198}{-113}\Rightarrow y=\dfrac{-9702}{113}\left(vô.lí\right)\\ \dfrac{z^2}{25}=\dfrac{198}{-113}\Rightarrow z^2=\dfrac{-4950}{113}\left(vô.lí\right)\)

Vậy ko có x,y,z thỏa mãn đề bài