a) Ta có: \(A=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)\cdot...\cdot\left(1-\dfrac{1}{2014}\right)\left(1-\dfrac{1}{2015}\right)\left(1-\dfrac{1}{2016}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2013}{2014}\cdot\dfrac{2014}{2015}\cdot\dfrac{2015}{2016}\)

\(=\dfrac{1}{2016}\)

b) Ta có: \(\dfrac{x-2}{12}+\dfrac{x-2}{20}+\dfrac{x-2}{30}+\dfrac{x-2}{42}+\dfrac{x-2}{56}+\dfrac{x-2}{72}=\dfrac{16}{9}\)

\(\Leftrightarrow\left(x-2\right)\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}\right)=\dfrac{16}{9}\)

\(\Leftrightarrow\left(x-2\right)\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}\right)=\dfrac{16}{9}\)

\(\Leftrightarrow\left(x-2\right)\left(\dfrac{1}{3}-\dfrac{1}{9}\right)=\dfrac{16}{9}\)

\(\Leftrightarrow\left(x-2\right)\cdot\dfrac{2}{9}=\dfrac{16}{9}\)

\(\Leftrightarrow x-2=\dfrac{16}{9}:\dfrac{2}{9}=\dfrac{16}{9}\cdot\dfrac{9}{2}=8\)

hay x=10

Vậy: x=10

\(\dfrac{x-1}{2016}+\dfrac{x-2}{2015}-\dfrac{x-3}{2014}=\dfrac{x-4}{2013}\)

\(\Leftrightarrow\dfrac{x-1}{2016}+\dfrac{x-2}{2015}=\dfrac{x-4}{2013}+\dfrac{x-3}{2014}\)

\(\Leftrightarrow\left(\dfrac{x-1}{2016}-1\right)+\left(\dfrac{x-2}{2015}-1\right)=\left(\dfrac{x-4}{2013}-1\right)+\left(\dfrac{x-3}{2014}-1\right)\)

\(\Leftrightarrow\dfrac{x-2017}{2016}+\dfrac{x-2017}{2015}=\dfrac{x-2017}{2013}+\dfrac{x-2017}{2014}\)

\(\Leftrightarrow\dfrac{x-2017}{2016}+\dfrac{x-2017}{2015}-\dfrac{x-2017}{2013}-\dfrac{x-2017}{2014}=0\)

\(\Leftrightarrow x-2017.\left(\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}-\dfrac{1}{2013}\right)=0\)

\(\text{Mà }\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}-\dfrac{1}{2103}\ne0\Rightarrow x-2017=0\)

\(\Leftrightarrow x=2017\)         \(\text{Vậy }x=2017\)

\(\dfrac{x+4}{2014}+\dfrac{x+3}{2015}=\dfrac{x+2}{2016}+\dfrac{x+1}{2017}\)

\(\dfrac{x+4}{2014}+1+\dfrac{x+3}{2015}+1=\dfrac{x+2}{2016}+1+\dfrac{x+1}{2017}+1\)

\(\dfrac{x+2018}{2014}+\dfrac{x+2018}{2015}=\dfrac{x+2018}{2016}+\dfrac{x+2018}{2017}\)

\(\left(x+2018\right)\left(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{2016}-\dfrac{1}{2017}\right)=0\\ x+2018=0\\ x=-2018\)

a) Tính chất dãy tỉ số bằng nhau: \(\dfrac{x+y}{2014}=\dfrac{x-y}{2016}=\dfrac{x+y+x-y}{2014+2016}=\dfrac{2x}{4030}=\dfrac{x}{2015}\)

\(\dfrac{x+y}{2014}=\dfrac{x-y}{2016}=\dfrac{x+y-x+y}{2014-2016}=\dfrac{2y}{-2}=\dfrac{y}{-1}\)

Nên: \(\dfrac{x}{2015}=\dfrac{y}{-1}=\dfrac{xy}{2015}\)

Xét: \(\left\{{}\begin{matrix}\dfrac{x}{2015}=\dfrac{xy}{2015}\Leftrightarrow2015x=2015xy\Leftrightarrow y=1\\\dfrac{y}{-1}=\dfrac{xy}{2015}\Leftrightarrow2015y=-1xy\Leftrightarrow2015=-1x\Leftrightarrow x=-2015\end{matrix}\right.\)

2) \(VT=\left|x-6\right|+\left|x-10\right|+\left|x-2022\right|+\left|y-2014\right|+\left|z-2015\right|\)

\(VT=\left|x-6\right|+\left|2022-x\right|+\left|x-10\right|+\left|y-2014\right|+\left|z-2015\right|\)

\(VT\ge\left|x-6+2022-x\right|+\left|x-10\right|+\left|y-2014\right|+\left|z-2015\right|\)

\(VT\ge2016+\left|x-10\right|+\left|y-2014\right|+\left|z-2015\right|\ge2016=VP\)

Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}6\le x\le2022\\x=10\\y=2014\\z=2015\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=10\\y=2014\\z=2015\end{matrix}\right.\)

\(=>\left(\dfrac{x+4}{2014}+1\right)+\left(\dfrac{x+3}{2015}+1\right)=\left(\dfrac{x+2}{2016}+1\right)+\left(\dfrac{x+1}{2017}+1\right)\)

=> \(\dfrac{x+2018}{2014}+\dfrac{x+2018}{2015}=\dfrac{x+2018}{2016}+\dfrac{x+2018}{2017}\)

=> (x+2018).\(\left(\dfrac{1}{2014}+\dfrac{1}{2015}\right)=\left(x+2018\right).\left(\dfrac{1}{2016}+\dfrac{1}{2017}\right)\)

=> (x+2018).\(\left(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{2016}-\dfrac{1}{2017}\right)\) = 0

Mà \(\dfrac{1}{2014}>0;\dfrac{1}{2015}>0;\dfrac{1}{2016}>0;\dfrac{1}{2017}>0\)

=>\(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{2016}-\dfrac{1}{2017}\ne0\)

=> \(x+2018=0\)

=>x = 0-2018

=> x = 0+(-2018)

=> x = -2018

\(\dfrac{x}{2012}+\dfrac{x+1}{2013}+\dfrac{x+2}{2014}+\dfrac{x+3}{2015}+\dfrac{x+4}{2016}=5\)

\(\Leftrightarrow\dfrac{x}{2012}+\dfrac{x+1}{2013}+\dfrac{x+2}{2014}+\dfrac{x+3}{2015}+\dfrac{x+4}{2016}-5=0\)

\(\Leftrightarrow\dfrac{x}{2012}-1+\dfrac{x+1}{2013}-1+\dfrac{x+2}{2014}-1+\dfrac{x+3}{2015}+\dfrac{x+4}{2016}-1=0\)

\(\Leftrightarrow\dfrac{x-2012}{2012}+\dfrac{x-2012}{2013}+\dfrac{x-2012}{2014}+\dfrac{x-2012}{2015}+\dfrac{x-2012}{2016}=0\)

\(\Leftrightarrow\left(x-12\right).\left(\dfrac{1}{2012}+\dfrac{1}{2013}+\dfrac{1}{2014}+\dfrac{1}{2015}+\dfrac{1}{2016}\right)=0\)

\(\Leftrightarrow x-12=0\)

\(\Leftrightarrow x=12\)