\(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}=\frac{x-y}{2-3}=\frac{y-z}{3-4}=\frac{x-z}{2-4}\) (T/c dãy tỷ số bằng nhau)

\(\Rightarrow\frac{x-z}{-2}=-\left(x-y\right)\left(1\right)\Rightarrow\frac{\left(x-z\right)^3}{-8}=-\left(x-y\right)^3=-\left(x-y\right)^2\left(x-y\right)\left(2\right)\)

\(\Rightarrow\frac{x-z}{-2}=-\left(y-z\right)\left(3\right)\)

Từ (1) và (3) \(\Rightarrow\left(x-y\right)=\left(y-z\right)\) Thay vào (2)

\(\Rightarrow\frac{\left(x-z\right)^3}{-8}=-\left(x-y\right)^2\left(y-z\right)\Rightarrow\left(x-z\right)^3=8\left(x-y\right)^2\left(y-z\right)\left(dpcm\right)\)

ta có 

\(A=\frac{3a-2b}{2a-3b}=\frac{\frac{3a}{b}-2}{\frac{2a}{b}-3}=\frac{\frac{3.5}{6}-2}{\frac{2.5}{6}-3}=\frac{\frac{1}{2}}{-\frac{4}{3}}=-\frac{3}{2}\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{18.19}+\frac{1}{19.20}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\)

\(=1-\frac{1}{20}=\frac{19}{20}\)

Vậy\(A=\frac{19}{20}\)

\(\frac{1}{2}a=\frac{2}{3}b\Rightarrow\frac{a}{\frac{2}{3}}=\frac{b}{\frac{1}{2}}\Rightarrow a=\frac{b}{\frac{3}{4}}\)

\(\Rightarrow\frac{a}{1}=\frac{b}{\frac{3}{4}}=\frac{c}{\frac{2}{3}}\)

Áp dụng tính chât dãy tỉ số bằng nhau ta có

\(\frac{a}{1}=\frac{b}{\frac{3}{4}}=\frac{c}{\frac{2}{3}}=\frac{a-b}{1-\frac{3}{4}}=\frac{15}{\frac{1}{4}}=60\)

\(\Rightarrow a=60.1=60\)

\(b=60.\frac{3}{4}=45\)

\(c=60.\frac{2}{3}=40\)

\(M=3\left|x-2013\right|+\left|x+1\right|+\left|x+2014\right|\)

hay \(M=\left|2013-x\right|+\left|x+1\right|+\left|2013-x\right|+\left|x+2014\right|+\left|2013-x\right|\)

mà ta có 

\(\hept{\begin{cases}\left|2013-x\right|\ge0\\\left|2013-x\right|+\left|x+1\right|\ge\left|2013-x+x+1\right|=2014\\\left|2013-x\right|+\left|x+2014\right|\ge\left|2013-x+x+2014\right|=4027\end{cases}}\)

Vậy \(M\ge2014+4027=6041\)

dấu bằng xay ra khi x=2013