\(\)\(\left|7x-5y\right|+ \left|2z-3x\right|=0\)
\(\left\{{}\begin{matrix}\left|7x-5y\right|\ge0\\\left|2z-3x\right|\ge0\end{matrix}\right.\)
\(\Rightarrow\left|7x-5y\right|+\left|2z-3x\right|\ge0\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\left|7x-5y\right|=0\Rightarrow7x=5y\\\left|2z-3x\right|=0\Rightarrow2z=3x\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{5}=\dfrac{y}{7}\\\dfrac{z}{3}=\dfrac{x}{2}\end{matrix}\right.\)
\(\Rightarrow\dfrac{x}{10}=\dfrac{y}{2}=\dfrac{z}{15}\)
Đặt:
\(\dfrac{x}{10}=\dfrac{y}{2}=\dfrac{z}{15}=k\)
\(\Rightarrow\left\{{}\begin{matrix}x=10k\\y=2k\\z=15k\end{matrix}\right.\)
Thay vào biểu thức ta có:(đã sửa đề)
\(10k.2k+2k.15k+10k.15k=2000\)
\(\Rightarrow20k^2+30k^2+150k^2=2000\)
\(\Rightarrow200k^2=2000\)
\(\Rightarrow k^2=10\Rightarrow k=\pm\sqrt{10}\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=10\sqrt{10}\\y=2\sqrt{10}\\z=15\sqrt{10}\end{matrix}\right.\\\left\{{}\begin{matrix}x=-10\sqrt{10}\\y=-2\sqrt{10}\\z=-15\sqrt{10}\end{matrix}\right.\end{matrix}\right.\)
\(\)
Ta có : \(\left|7x-5y\right|+\left|2z-3x\right|=0\)
=> \(\left[{}\begin{matrix}7x-5y=0\\2z-3x=0\end{matrix}\right.=>\left[{}\begin{matrix}7x=5y\\2z=3x\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}\dfrac{x}{5}=\dfrac{y}{7}\\\dfrac{z}{3}=\dfrac{x}{2}\end{matrix}\right.=>\left[{}\begin{matrix}\dfrac{x}{10}=\dfrac{y}{14}\\\dfrac{z}{15}=\dfrac{x}{10}\end{matrix}\right.=>\dfrac{x}{10}=\dfrac{y}{14}=\dfrac{z}{15}\)
P/S : tự làm nốt nha
