Question

x:y:z=3:4:5 và 2x\(^2\)+2y\(^2\)-3z\(^2\)=100

Áp dụng t/c của dãy tỉ số bằng nhau

Answer 1

Đặt \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=3k\\y=4k\\z=5k\end{matrix}\right.\)

Ta có: \(2x^2+2y^2-3z^2=100\)

\(\Leftrightarrow2\cdot9k^2+2\cdot16k^2-3\cdot25k^2=100\)

\(\Leftrightarrow-25k^2=100\)(vô lý)

Answer 2

\(x:y:z=3:4:5\\ \Rightarrow\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=k\\ \Rightarrow\left\{{}\begin{matrix}x=3k\\y=4k\\z=5k\end{matrix}\right.\)

Ta có \(2x^2+2y^2-3z^2=100\)

\(\Rightarrow2.\left(3k\right)^2+2\left(4k\right)^2-3.\left(5k\right)^2=100\\ \Rightarrow18k^2+32k^2-75k^2=100\\ \Rightarrow\left(18+32-75\right)k^2=100\\ \Rightarrow k^2=-4\left(Vôlý\right)\)