2019 + 2017 + 2015 + .... + 1003 + 1001 - 2018 - 2016 - 2014 -.... - 1002 - 1000

=(2019-2018)+(2017-2016)+....+(1001-1000)

=1+1+...+1

= 510

S = 2020 + 2019 - 2018 - 2017 + 2016 + 2015 - 2014 - 2013 + ... + 4 + 3 - 2 - 1

= ( 2020 + 2019 - 2018 - 2017 ) + ( 2016 + 2015 - 2014 - 2013 ) + ... + ( 4 + 3 - 2 - 1 )   (có tất cả 2020 : 4 = 505 nhóm)

= 4 + 4 + ... + 4

= 4. 505 = 2020

Vậy S = 2020.

S= 2020

Bạn huyền đúng rồi đó .

hok tốt

\(\frac{2015+2016\times2017}{2017\times2018-2019}\)

\(=\frac{2015+4066272}{4070306-2019}\)

\(=\frac{4068287}{4068287}\)

\(=1\)

cảm ơn bạn

Ta có: \(\frac{x-2019}{2018}+\frac{x-2018}{2017}=\frac{x-2017}{2016}+\frac{x-2016}{2015}\)

\(\Leftrightarrow\left(\frac{x-2019}{2018}+1\right)+\left(\frac{x-2018}{2017}+1\right)=\left(\frac{x-2017}{2016}+1\right)+\left(\frac{x-2016}{2015}+1\right)\)

\(\Leftrightarrow\frac{x-1}{2018}+\frac{x-1}{2017}=\frac{x-1}{2016}+\frac{x-1}{2015}\)

\(\Leftrightarrow\frac{x-1}{2018}+\frac{x-1}{2017}-\frac{x-1}{2016}-\frac{x-1}{2015}=0\)

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

\(\Leftrightarrow x-1=0\)( vì \(\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2016}-\frac{1}{2015}\ne0\))

\(\Leftrightarrow x=1\)

Vạy x=1

xin lỗi bạn viết rõ ra đc o

\(2015+2016.2017\)

\(2015+2015+1.2016+1\)

\(1+2016+1\)

\(1+2016=2017\)

\(2017.2018-2019=1\)

\(2018-1.2019-1.2020-1\)

\(1.\left(2020-2019\right).1.\left(2018-2017\right)\)

\(1.1=1\)

Bài làm:

(2019-2018+2017-.....-2) x (100 -25x2x2)

=(2019-2018+2017-.....-2) x (100 -25x4)

=(2019-2018+2017-.....-2) x 0

=0

*like phát

=(2019 – 2018 + 2017 – 2016 + 2015 + ....... – 4 + 3 – 2) x(100-25x4)

=(2019 – 2018 + 2017 – 2016 + 2015 + ....... – 4 + 3 – 2) x(100-100)

=(2019 – 2018 + 2017 – 2016 + 2015 + ....... – 4 + 3 – 2) x0

=0

\(\frac{2016\cdot2017+2015}{2018\cdot2017-2019}\)

= \(\frac{2016\cdot2017+2015}{\left(2016+2\right)\cdot2017-2019}\)

= \(\frac{2016\cdot2017+2015}{2016\cdot2017+2\cdot2017-2019}\)

= \(\frac{2016\cdot2017+2015}{2016\cdot2017+4034-2019}\)

= \(\frac{2016\cdot2017+2015}{2016\cdot2017-2015}\)

= \(1\)

\(\frac{2016\times2017+2015}{2018\times2017-2019}\)

\(=\frac{2016\times2017+2015}{\left(2016+2\right)\times2017-2019}\)

\(=\frac{2016\times2017+2015}{2016\times2017+2\times2017-2019}\)

\(=\frac{2016\times2017+2015}{2016\times2017+4034-2019}\)

\(=\frac{2016\times2017+2015}{2016\times2017+2015}\)

\(=1\)

#)Giải :

\(Q=2+\frac{2016}{2017+2018+2019}+\frac{2017}{2017+2018+2019}+\frac{2018}{2017+2018+2019}\)

Ta thấy : \(2>\frac{2016}{2017};2>\frac{2017}{2018};2>\frac{2018}{2019}\left(1\right)\)

\(\frac{2016}{2017+2018+2019}< \frac{2016}{2017}\left(2\right)\)

\(\frac{2017}{2017+2018+2019}< \frac{2017}{2018}\left(3\right)\)

\(\frac{2018}{2017+2018+2019}< \frac{2018}{2019}\left(4\right)\)

Từ (1) (2) (3) (4) \(\Rightarrow P>Q\)