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Ta có : \(2018.\left(\frac{1}{2017}-\frac{2019}{1009}\right)-2019.\left(\frac{1}{2017}-2\right)=\frac{2018}{2017}-2019.2-\frac{2019}{2017}+2019.2\)
\(=\frac{2018}{2017}-\frac{2019}{2017}=-\frac{1}{2017}\)
\(2018.\left(\frac{1}{2017}-\frac{2019}{1009}\right)-2019.\left(\frac{1}{2017}-2\right)\)
\(=\frac{2018}{2017}-2018.\frac{2019}{1009}-\frac{2019}{2017}+2019.2\)
\(=\frac{2018}{2017}-2.2019-\frac{2019}{2017}+2.2019\)
\(=\frac{2018}{2017}-\frac{2019}{2017}=-\frac{1}{2017}\)
\(\frac{x-4}{2017}+\frac{x-3}{2018}+\frac{x-2}{2019}+\frac{x-1}{2020}=4\\ \Leftrightarrow\left(\frac{x-4}{2017}-1\right)+\left(\frac{x-3}{2018}-1\right)+\left(\frac{x-2}{2019}-1\right)+\left(\frac{x-1}{2020}-1\right)=4-1-1-1\)
\(\Leftrightarrow\frac{x-2021}{2017}+\frac{x-2021}{2018}+\frac{x-2021}{2019}+\frac{x-2021}{2020}=0\)
\(\Leftrightarrow\left(x-2021\right)\left(\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}+\frac{1}{2020}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-2021=0\\\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}+\frac{1}{2020}\ne0\end{matrix}\right.\)
\(\Leftrightarrow x=2021\)
Vậy...
\(A=\frac{\left|x-2017\right|+2018}{\left|x-2017\right|+2019}\)
\(A=\frac{\left|x-2017\right|+2019-1}{\left|x-2017\right|+2019}\)
\(A=1-\frac{1}{\left|x-2017\right|+2019}\)
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
\(M=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2019}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{2017}+\frac{1}{2019}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)
\(=\left(1+\frac{1}{2}+...+\frac{1}{2018}+\frac{1}{2019}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)
\(=\left(1+\frac{1}{2}+...+\frac{1}{2018}+\frac{1}{2019}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1009}\right)\)
\(=\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2019}=N\)
\(\Rightarrow M-N=0\Rightarrow\left(M-N\right)^2=0\)
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)
\(2018\cdot\left(\frac{1}{2017}-\frac{2019}{1009}\right)-2019\cdot\left(\frac{1}{2017}-2\right)=\frac{2018}{2017}-4038-\frac{2019}{2017}+4038\)
\(=\frac{2018}{2017}-\frac{2019}{2017}=-\frac{1}{2017}\)