\(2018\cdot2018-2017\cdot2019\)

\(=2018\cdot2018-2017\left(2018-1\right)\)

\(=2018\cdot2018-2017\cdot2018-2017\cdot1\)

\(=2018\left(2018-2017\right)-2017\cdot1\)

\(=2018\cdot1-2017\cdot1\)

\(=1\)

\(2018\cdot2018-2017\cdot2019\)

\(=(2019-2018)\cdot(2018-2017)\)

\(=1\)

\(2018\times2018-\left(2018-1\right)\times\left(2018+1\right)=2018\times2018-2018\times2018-2018+2018-1\)

\(=-1\)

các bạn hãy trả lời nhanh cho mình với nhé

a) 2018 x 2019 - 2017  = 2017 = 1                                                        b) 2018 x 2018            = 2018 x 2018 = 2018

     2017 x 2018  + 2019   2017                                                                  (2017 + 1) x 2019       2018 x 2019     2019

chúc bạn hok tốt

bạn sky ơi! Bài a) bạn có thể giải thích rõ cho mình vì sao bằng 1 dc ko?

Gọi 1/4 số a là 0,25 . Ta có :

                   a . 3 - a . 0,25 = 147,07

                   a . (3 - 0,25) = 147,07 ( 1 số nhân 1 hiệu )

                      a . 2,75 = 147,07

                         a = 147,07 : 2,75

                          a = 53,48

< chắc cắn luôn

Ta có :

\(A=2016.2020\) 

\(\Rightarrow A=\left(2018-2\right)\left(2018+2\right)\)

\(\Rightarrow A=2018^2+2.2018-2.2018-4\) (1)

\(\Rightarrow A=2018^2-4\)

\(B=2018.2018\)

\(\Rightarrow B=2018^2\) (2)

Từ (1) và (2) ta có :

\(2018^2-4< 2018^2\Rightarrow A< B\)

k đi chứ bn!

= 1 nhé bạn

\(\frac{2016+2017.2018}{2017.2019-1}\)

= \(\frac{2016+2017.2018}{2017.2018+2017-1}\)

= \(\frac{2016+2017.2018}{2017.2018+2016}\)

= 1

\(B=\dfrac{2017\times2018+1000}{2017\times2018+2018-1018}\\ B=\dfrac{2017\times2018+1000}{2017\times2018+1000}\\ B=1\)

B = 3

B = 1

\(\frac{2016+2017.2018}{2017.2019-1}\)

\(=\frac{\left(2016+1\right)+2017.2018-1}{2017.2019-1}\)

\(=\frac{2017+2017.2018-1}{2017.2019-1}\)

\(=\frac{2017.\left(1+2018\right)-1}{2017.2019-1}\)

\(=\frac{2017.2019-1}{2017.2019-1}=1\)

\(\frac{2016+2017\times2018}{2017\times2019-1}\)

\(=\frac{2016+2017\times2018}{2017\times\left(2018+1\right)-1}\)

\(=\frac{2016+2017\times2018}{2017\times2018+2017-1}\)

\(=\frac{2016+2017\times2018}{2017\times2018+2016}\)

\(=1\)

__CHÚC BN HOK TỐT__

\(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{2017.2019}\)

\(=\frac{3}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2017}-\frac{1}{2019}\right)\)

\(=\frac{3}{2}.\left(1-\frac{1}{2019}\right)\)

\(=\frac{3}{2}.\frac{2018}{2019}\)

\(=\frac{1009}{673}\)

\(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}.....+\frac{3}{2017.2019}\)

\(=\frac{3}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{2017.2019}\right)\)

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

\(=\frac{3}{2}\left(1-\frac{1}{2019}\right)\)

\(=\frac{3}{2}.\frac{2018}{2019}=\frac{1009}{673}\)

\(\frac{3}{1\times3}+\frac{3}{3\times5}+.......+\frac{3}{2017\times2019}\)

\(=\frac{3}{2}\times\left(\frac{2}{1\times3}+\frac{2}{3\times5}+......+\frac{2}{2017\times2019}\right)\)

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

\(=\frac{3}{2}\times\left(1-\frac{1}{2019}\right)\)

\(=\frac{3}{2}\times\frac{2018}{2019}\)

\(=\frac{1009}{673}\)

           p=1/(3*5)+1/(5*7)+.....+1/(2015*2017)+1/(2017*2019)

<=> p = 1/3-1/5+1/5-1/7+1/7-......+1/2017-1/2019

<=> p = 1/3 - 1/2019

<=> p = 224/673

\(P=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{2015.2017}+\frac{1}{2017.2019}\)

\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{2017}-\frac{1}{2019}\right)\)

\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{2019}\right)\)

\(=\frac{112}{673}\)

\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{2015.2017}+\frac{1}{2017.2019}\)

\(=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+....+\frac{2}{2015.2017}+\frac{2}{2017.2019}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{2015}-\frac{1}{2017}+\frac{1}{2017}-\frac{1}{2019}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{2019}\right)=\frac{1}{2}.\frac{224}{673}=\frac{112}{673}\)