Question

[Panda Cute]

\(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}v\text{à}2x^2+2y^2-3z^2=-100\)

Tìm x,y,z

Answer 1

[Mai Tiến Đỗ]

\(2x^2+2y^2-3z^2=-100\left(1\right)\)

Đặt \(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=k\)\(\Rightarrow\left\{{}\begin{matrix}x=3k\\y=4k\\z=5k\end{matrix}\right.\)\(\left(2\right)\)

Thay (2) vào (1) ta được:

\(2.\left(3k\right)^2+2.\left(4k\right)^2-3.\left(5k\right)^2=-100\)

\(\Leftrightarrow18k^2+32k^2-75k^2=-100\)

\(\Leftrightarrow-25k^2=-100\)

\(\Leftrightarrow k^2=4\)

\(\Leftrightarrow k=\pm2\)

TH1: Thay k=2 vào (2) ta được

\(\left\{{}\begin{matrix}x=3.2=6\\y=4.2=8\\z=5.2=10\end{matrix}\right.\)

TH2: Thay k=-2 vào (2) ta được:

\(\left\{{}\begin{matrix}x=3.\left(-2\right)=-6\\y=4.\left(-2\right)=-8\\z=5.2\left(-2\right)=-10\end{matrix}\right.\)

Vậy \(\left(x,y,z\right)=\left\{\left(6,8,10\right);\left(-6,-8,-10\right)\right\}\)

Answer 2

[Vũ Minh Tuấn]

Ta có: \(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}.\)

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

=> \(\frac{2x^2}{18}=\frac{2y^2}{32}=\frac{3z^2}{75}\) và \(2x^2+2y^2-3z^2=-100.\)

Áp dụng tính chất dãy tỉ số bằng nhau ta được:

\(\frac{2x^2}{18}=\frac{2y^2}{32}=\frac{3z^2}{75}=\frac{2x^2+2y^2-3z^2}{18+32-75}=\frac{-100}{-25}=4.\)

\(\Rightarrow\left\{{}\begin{matrix}\frac{x^2}{9}=4\Rightarrow x^2=36\Rightarrow\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\\\frac{y^2}{16}=4\Rightarrow y^2=64\Rightarrow\left[{}\begin{matrix}y=8\\y=-8\end{matrix}\right.\\\frac{z^2}{25}=4\Rightarrow z^2=100\Rightarrow\left[{}\begin{matrix}z=10\\z=-10\end{matrix}\right.\end{matrix}\right.\)

Vậy \(\left(x;y;z\right)=\left(6;8;10\right),\left(-6;-8;-10\right).\)

Chúc bạn học tốt!