Đáp án:

$ \left[\begin{array}{l} \left\{\begin{array}{l} x=15\\ y=10 \\ z =6\end{array} \right.\\ \left\{\begin{array}{l} x=-15\\ y=-10 \\ z =-6\end{array} \right.\end{array} \right.$

Giải thích các bước giải:

Đặt $2x = 3y = 5z=k$

$\Rightarrow \left\{\begin{array}{l} x=\dfrac{k}{2}\\ y=\dfrac{k}{3} \\ z =\dfrac{k}{5}\end{array} \right.\\ xy + yz + zx = 300\\ \Leftrightarrow \dfrac{k}{2}.\dfrac{k}{3}+\dfrac{k}{3}.\dfrac{k}{5}+\dfrac{k}{5}.\dfrac{k}{2}=300\\ \Leftrightarrow \dfrac{k^2}{6}+\dfrac{k^2}{15}+\dfrac{k^2}{10}=300\\ \Leftrightarrow \dfrac{5k^2}{30}+\dfrac{2k^2}{30}+\dfrac{3k^2}{30}=300\\ \Leftrightarrow \dfrac{10k^2}{30}=300\\ \Leftrightarrow 10k^2=9000\\ \Leftrightarrow k^2=900\\ \Leftrightarrow k=\pm 30\\ \Rightarrow \left[\begin{array}{l} \left\{\begin{array}{l} x=\dfrac{k}{2}=15\\ y=\dfrac{k}{3}=10 \\ z =\dfrac{k}{5}=6\end{array} \right.\\ \left\{\begin{array}{l} x=\dfrac{k}{2}=-15\\ y=\dfrac{k}{3}=-10 \\ z =\dfrac{k}{5}=-6\end{array} \right.\end{array} \right.$

$\\$

Có : `2x=3y=5z`

`-> (2x)/30=(3y)/30=(5z)/30`

`-> x/15=y/10=z/6`

Đặt `x/15=y/10=z/6=k (k \ne 0)`

`->` $\begin{cases} \dfrac{x}{15}=k\\\dfrac{y}{10}=k\\\dfrac{z}{6}=k \end{cases}$ `->` $\begin{cases} x=15k\\y=10k\\z=6k \end{cases}$

Có : `xy + yz + xz = 300`

`-> 15k . 10k + 10k . 6k + 15k . 6k =300`

`-> (15 . 10) (k.k) + (10.6) (k.k) + (15 . 6) (k.k)=300`

`-> 150k^2 + 60k^2 + 90k^2=300`

`-> (150 + 60+90)k^2=300`

`-> 300k^2=300`

`-> k^2=300 : 300`

`-> k^2=1`

`->` \(\left[ \begin{array}{l}k^2=1^2\\k^2=(-1)^2\end{array} \right.\) `->` \(\left[ \begin{array}{l}k=1\\k=-1\end{array} \right.\) (Thỏa mãn)

Với `k=1`

`->` $\begin{cases} x=15.1\\y=10.1\\z=6.1 \end{cases}$ `->` $\begin{cases} x=15\\y=10\\z=6 \end{cases}$

Với `k=-1`

`->` $\begin{cases} x=15.(-1)\\y=10.(-1)\\z=6.(-1) \end{cases}$ `->` $\begin{cases} x=-15\\y=-10\\z=-6 \end{cases}$

Vậy `(x;y;z) = (15;10;6), (-15;-10;-6)`