\(\frac{x}{2}=\frac{y}{3}\Leftrightarrow\frac{x}{8}=\frac{y}{12}\)

\(\frac{y}{4}=\frac{z}{5}\Leftrightarrow\frac{y}{12}=\frac{z}{15}\)

suy ra \(\frac{x}{8}=\frac{y}{12}=\frac{z}{15}\).

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

\(\frac{x}{8}=\frac{y}{12}=\frac{z}{15}=\frac{x+y+z}{8+12+15}=\frac{10}{35}=\frac{2}{7}\)

\(\Leftrightarrow\hept{\begin{cases}x=\frac{2}{7}.8=\frac{16}{7}\\y=\frac{2}{7}.12=\frac{24}{7}\\z=\frac{2}{7}.15=\frac{30}{7}\end{cases}}\)

ta có : \(\frac{x}{7}=\frac{y}{4}=\frac{z}{2}\) =  \(\frac{3z}{6}\)

theo tính chất của dãy tỉ số ta có : 

\(\frac{x}{7}=\frac{y}{4}=\frac{3z}{6}\)= \(\frac{x-3z}{7-6}\)= \(\frac{9}{1}=9\)

\(\frac{x}{7}=9\Rightarrow x=9\times7=63\)

\(\frac{y}{4}=9\Rightarrow y=9\times4=36\)

\(\frac{z}{2}=9\Rightarrow z=9\times2=18\)

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Bài 4. 

\(\left|x-1\right|+\left|y-2\right|+\left(z-x\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}x-1=0\\y-2=0\\z-x=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=z=1\\y=2\end{cases}}\)

Bài 3. 

\(\left|x-1\right|+\left|2x-2\right|+\left|4x-4\right|+\left|5x-5\right|=36\)

\(\Leftrightarrow\left|x-1\right|+2\left|x-1\right|+4\left|x-1\right|+5\left|x-1\right|=36\)

\(\Leftrightarrow12\left|x-1\right|=36\)

\(\Leftrightarrow\left|x-1\right|=3\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=3\\x-1=-3\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=-2\end{cases}}\)