có ai giúp với

\(\text{Đặt x=4y}=\frac{z-9}{125}=k\)

\(\Leftrightarrow\hept{\begin{cases}x=k\\y=\frac{1}{4}k\\z=125k+9\end{cases}}\)

\(\Rightarrow\frac{505}{4}k=2029\)

\(\text{Mà x+y+z=2029}\)

\(\Rightarrow k+\frac{1}{4}k+125k+9=2029\)

\(\Rightarrow\frac{505}{4}k=2020\Rightarrow k=16\)

\(\Rightarrow\hept{\begin{cases}x=16\\y=4\\z=2009\end{cases}}\)

Giải:
Đặt \(x=4y=\dfrac{z-9}{125}=k\Rightarrow\left\{{}\begin{matrix}x=k\\y=\dfrac{1}{4}k\\z=125k+9\end{matrix}\right.\)

Mà \(x+y+z=2029\)

\(\Rightarrow k+\dfrac{1}{4}k+125k+9=2029\)

\(\Rightarrow\dfrac{505}{4}k=2020\)

\(\Rightarrow k=16\)

\(\Rightarrow\left\{{}\begin{matrix}x=16\\y=4\\z=2009\end{matrix}\right.\)

Vậy \(x=16;y=4;z=2009\)

\(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\le\frac{1}{4x}+\frac{1}{4y}+\frac{1}{4z}+\frac{9}{4}\)

\(\Leftrightarrow\left(x+y+z\right)\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)\le\left(x+y+z\right)\left(\frac{1}{4x}+\frac{1}{4y}+\frac{1}{4z}\right)+\frac{9}{4}\)

\(\Leftrightarrow\frac{z}{x+y}+\frac{x}{y+z}+\frac{y}{z+x}\le\frac{y+z}{4x}+\frac{z+x}{4y}+\frac{x+y}{4z}\)

Ta có:

\(VP=\frac{1}{4}\left(\frac{x}{y}+\frac{x}{z}+\frac{y}{x}+\frac{y}{z}+\frac{z}{x}+\frac{z}{y}\right)\)

\(\ge\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}=VT\)

1) \(\frac{x}{2}=\frac{y}{3}=\frac{z}{7}=\frac{2x-4y+3z}{2.2-4.3+3.7}=\frac{-39}{13}=-3\)

\(\Leftrightarrow\hept{\begin{cases}x=-3.2=-6\\y=-3.3=-9\\z=-3.7=-21\end{cases}}\)

2) \(9x=10y\Leftrightarrow\frac{x}{10}=\frac{y}{9},4y=3z\Leftrightarrow\frac{y}{9}=\frac{z}{12}\)

suy ra \(\frac{x}{10}=\frac{y}{9}=\frac{z}{12}=\frac{x-y+z}{10-9+12}=\frac{78}{13}=6\)

\(\Leftrightarrow\hept{\begin{cases}x=6.10=60\\y=6.9=54\\z=6.12=72\end{cases}}\)

3) \(3x=4y=6z\Leftrightarrow\frac{x}{4}=\frac{y}{3}=\frac{z}{2}=\frac{x-y+z}{4-3+2}=\frac{-9}{3}=-3\)

\(\Leftrightarrow\hept{\begin{cases}x=-3.4=-12\\y=-3.3=-9\\z=-3.2=-6\end{cases}}\)