\(\left\{{}\begin{matrix}3\left(x-1\right)=2\left(y-2\right)\\5\left(y-2\right)=4\left(z-3\right)\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3\left(x-1\right)}{6}=\dfrac{2\left(y-2\right)}{6}\\\dfrac{5\left(y-2\right)}{20}=\dfrac{4\left(z-3\right)}{20}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x-1}{2}=\dfrac{y-2}{3}\\\dfrac{y-2}{4}=\dfrac{z-3}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x-1}{8}=\dfrac{y-2}{12}\\\dfrac{y-2}{12}=\dfrac{z-3}{15}\end{matrix}\right.\Leftrightarrow\dfrac{x-1}{8}=\dfrac{y-2}{12}=\dfrac{z-3}{15}\Leftrightarrow\dfrac{2x-2}{16}=\dfrac{3y-6}{36}=\dfrac{z-3}{15}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{2x-2}{16}=\dfrac{3y-6}{36}=\dfrac{z-3}{15}=\dfrac{2x-2+3y-6-z+3}{16+36-15}=\dfrac{\left(2x+3y-z\right)+\left(3-2-6\right)}{37}=\dfrac{79-5}{37}=\dfrac{74}{37}=2\)
\(\Rightarrow\left\{{}\begin{matrix}x=2.8+1=17\\y=2.12+2=26\\z=2.15+3=33\end{matrix}\right.\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{3a+b+c}{a}=\dfrac{a+3b+c}{b}=\dfrac{a+b+3c}{c}=\dfrac{3a+b+c+a+3b+c+a+b+3c}{a+b+c}=\dfrac{5a+5b+5c}{a+b+c}=\dfrac{5\left(a+b+c\right)}{a+b+c}=5\)\(\Rightarrow\left\{{}\begin{matrix}3a+b+c=5a\\a+3b+c=5b\\a+b+3c=5c\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b+c=2a\\a+c=2b\\a+b=2c\end{matrix}\right.\)
\(M=\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}=\dfrac{2c}{c}+\dfrac{2a}{a}+\dfrac{2b}{b}=2+2+2=6\)