\(2018\cdot\left(\frac{1}{2017}-\frac{2019}{1009}\right)-2019\cdot\left(\frac{1}{2017}-2\right)=\frac{2018}{2017}-4038-\frac{2019}{2017}+4038\)

\(=\frac{2018}{2017}-\frac{2019}{2017}=-\frac{1}{2017}\)

Ta có : \(2018.\left(\frac{1}{2017}-\frac{2019}{1009}\right)-2019.\left(\frac{1}{2017}-2\right)=\frac{2018}{2017}-2019.2-\frac{2019}{2017}+2019.2\)

\(=\frac{2018}{2017}-\frac{2019}{2017}=-\frac{1}{2017}\)

\(2018.\left(\frac{1}{2017}-\frac{2019}{1009}\right)-2019.\left(\frac{1}{2017}-2\right)\)

\(=\frac{2018}{2017}-2018.\frac{2019}{1009}-\frac{2019}{2017}+2019.2\)

\(=\frac{2018}{2017}-2.2019-\frac{2019}{2017}+2.2019\)

\(=\frac{2018}{2017}-\frac{2019}{2017}=-\frac{1}{2017}\)

help me

\(a)\) Ta có : 

\(VP=\frac{2018}{1}+\frac{2017}{2}+\frac{2016}{3}+...+\frac{2}{2017}+\frac{1}{2018}\)

\(VP=\left(\frac{2018}{1}-1-...-1\right)+\left(\frac{2017}{2}+1\right)+\left(\frac{2016}{3}+1\right)+...+\left(\frac{2}{2017}+1\right)+\left(\frac{1}{2018}+1\right)\)

\(VP=1+\frac{2019}{2}+\frac{2019}{3}+...+\frac{2019}{2017}+\frac{2019}{2018}\)

\(VP=2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}\right)\)

Lại có : 

\(VT=\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\right).x\)

\(\Rightarrow\)\(x=2019\)

Vậy \(x=2019\)

Chúc bạn học tốt ~ 

\(b)\) \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2017}{2019}\)

\(\Leftrightarrow\)\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2017}{2019}\)

\(\Leftrightarrow\)\(2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2017}{2019}\)

\(\Leftrightarrow\)\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2017}{2019}\)

\(\Leftrightarrow\)\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2017}{2019}\)

\(\Leftrightarrow\)\(2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2017}{2019}\)

\(\Leftrightarrow\)\(1-\frac{2}{x+1}=\frac{2017}{2019}\)

\(\Leftrightarrow\)\(\frac{2}{x+1}=1-\frac{2017}{2019}\)

\(\Leftrightarrow\)\(\frac{2}{x+1}=\frac{2}{2019}\)

\(\Leftrightarrow\)\(x+1=2019\)

\(\Leftrightarrow\)\(x=2019-1\)

\(\Leftrightarrow\)\(x=2018\)

Vậy \(x=2018\)

Chúc bạn học tốt ~ 

Đề có sai ko bạn sao lại c-d ?

Sửa đề : Cần chứng minh \(4\left(a-b\right)\left(b-c\right)=\left(c-a\right)^2\)

Đặt :\(\frac{a}{2017}=\frac{b}{2018}=\frac{c}{2019}=k\)

\(\Rightarrow\hept{\begin{cases}a=2017k\\b=2018k\\c=2019k\end{cases}}\)

Khi đó :

\(4\left(a-b\right)\left(b-c\right)=4\left(2017k-2018k\right)\left(208k-2019k\right)\)

\(=4\cdot\left(-k\right)\cdot\left(-k\right)=4k^2\)

\(\left(c-a\right)^2=\left(2019k-2017k\right)^2=\left(2k\right)^2=4k^2\)

Do đó : \(4\left(a-b\right)\left(b-c\right)=\left(c-a\right)^2\) (đpcm)

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)

\(\Leftrightarrow\frac{1}{x}+\frac{1}{y}=\frac{1}{x+y+z}-\frac{1}{z}\)

\(\Leftrightarrow\frac{x+y}{xy}=\frac{z}{\left(x+y+z\right).z}-\frac{x+y+z}{z.\left(x+y+z\right)}=\frac{-x-y}{z.\left(x+y+z\right)}\)

\(\Leftrightarrow\frac{x+y}{xy}=\frac{x+y}{-z.\left(x+y+z\right)}\)

TH1: x+y=0

=> x=-y => P=0

TH2: xy=-z.(x+y+z)

\(\Leftrightarrow xy=-xz-zy-z^2\Leftrightarrow xy+xz+zy+z^2=0\Leftrightarrow x.\left(y+z\right)+z.\left(y+z\right)=0\)

\(\Leftrightarrow\left(x+z\right).\left(y+z\right)=0\Leftrightarrow\orbr{\begin{cases}x=-z\\y=-z\end{cases}\Rightarrow P=0}\)

\(A=\frac{\left|x-2017\right|+2018}{\left|x-2017\right|+2019}\)

\(A=\frac{\left|x-2017\right|+2019-1}{\left|x-2017\right|+2019}\)

\(A=1-\frac{1}{\left|x-2017\right|+2019}\)

A nhỏ nhất khi \(1-\frac{1}{\left|x-2017\right|+2019}\)nhỏ nhất

khi \(\frac{1}{\left|x-2017\right|+2019}\)lớn nhất

khi \(\left|x-2017\right|+2019\)nhỏ nhất

mà |x - 2017| \(\ge0\)

=> |x - 2017| + 2019 \(\ge2019\)

Vậy A nhỏ nhất khi A = 2019 khi x - 2017 = 0 => x = 2017

\(A=\frac{\backslash x-2017\backslash+2018}{\backslash x-2017\backslash+2019}\) 

\(A=\frac{2018}{2019}\)

Ta có : \(A=\frac{\left|x-2017\right|+2018}{\left|x-2017\right|+2019}=\frac{\left|x-2017\right|+2019-1}{\left|x-2017\right|+2019}\)

\(=1-\frac{1}{\left|x-2017\right|+2019}\)

Ta có : \(\left|x-2017\right|\ge0\)

\(\Rightarrow\left|x-2017\right|+2019\ge2019\)

\(\Rightarrow\frac{1}{\left|x-2017\right|+2019}\le\frac{1}{2019}\)

\(\Rightarrow-\frac{1}{\left|x-2017\right|+2019}\ge-\frac{1}{2019}\)

\(\Rightarrow1-\frac{1}{\left|x-2017\right|+2019}\ge1-\frac{1}{2019}=\frac{2018}{2019}\)

Hay : \(A\ge\frac{2018}{2019}\)

Dấu "=" xảy ra \(\Leftrightarrow\left|x-2017\right|=0\Leftrightarrow x=2017\)

Vậy : min \(A=\frac{2018}{2019}\) tại \(x=2017\)