= 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

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?

           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

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

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

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

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

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

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

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

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

Ta có:

\(A=\frac{2017\cdot2018-1}{2017\cdot2018-2}\)

\(A=\frac{2017\cdot2018-2+1}{2017\cdot2018-2}\)

\(A=\frac{2017\cdot2018-2}{2017\cdot2018-2}+\frac{1}{2017\cdot2018-2}\)

\(A=1+\frac{1}{2017\cdot2018-2}\)

Ta có phân số trung gian là 1. Ta có:
\(A>1\) ; \(B< 1\)

\(\Rightarrow A>1>B\)

\(\Rightarrow A>B\)

Vậy A>B
Chúc em học tốt!

\(\Rightarrow\text{❤️✔✨♕✨✔️❤ }\Leftarrow\)

\(\text{Ta có :}\)

\(A=\frac{2017\cdot2018-1}{2017\cdot2018-2}=\frac{4070305}{4070304}=1\frac{1}{4070304}\)

\(B=\frac{2017}{2018}\)

\(\text{Vì : }1\frac{1}{4070304}>1\text{ mà }\frac{2017}{2018}< 1\text{ nên }1\frac{1}{4070304}>\frac{2017}{2018}\)

\(\Rightarrow A>B\)

Ta có :

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

= ( 2017 + 1 ) x 2019 - ( 2017 + 1 ) 

= 2017 x 2019 + 2019 - 2017 - 1

= 2017 x 2019 + 1 > 2017 x 2019

\(\Rightarrow\frac{2018\times2018}{2017\times2019}=\frac{2017\times2019+1}{2017\times2019}=1+\frac{1}{2017\times2019}>1\)

Vậy ta chọn B

~~Học tốt~~

các bn chọ hộ mình, đúng mình k cho

\(C=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)

\(C=1-\frac{1}{2018}\)

\(C=\frac{2017}{2018}\)

\(C=\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+.....+\frac{1}{2017x2018}\)

Ta thấy \(\frac{1}{1x2}=\frac{1}{1}-\frac{1}{2}\)

               \(\frac{1}{2x3}=\frac{1}{2}-\frac{1}{3}\)

      .............................................

           \(\frac{1}{2017x2018}=\frac{1}{2017}-\frac{1}{2018}\)

\(\Rightarrow C=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{2017}-\frac{1}{2018}\)

\(\Rightarrow C=\frac{1}{1}-\frac{1}{2018}\)

\(\Rightarrow C=\frac{2017}{2018}\)

Chúc bạn học tốt nhớ k mình nhá

\(C=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}=1-\frac{1}{2018}=\frac{2017}{2018}\)

\(\frac{2015}{2016}\)+ \(\frac{1}{2016}\)x 2017 +\(\frac{1}{2017}\) x2018 =\(\frac{6052}{2017}\)

\(1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\)

\(=1+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\right)\)

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

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

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

\(=\frac{2018}{2018}+\frac{2017}{2018}=\frac{4035}{2018}\)

\(1+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}...+\frac{1}{2017\cdot2018}\)

\(=1+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}...+\frac{1}{2017}-\frac{1}{2018}\right)\)

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

\(=1+\frac{2017}{2018}\)

\(=1+\frac{2017}{2018}\)

\(=\frac{4035}{2018}\)

\(1+\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{2017x2018}\)

\(=1+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\right)\)

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

\(=1+\frac{2017}{2018}\)

\(=\frac{4035}{2018}\)

Ta có :

\(\frac{2017\times2018+1}{2019+2016\times2018}\)

\(=\frac{2017\times2018+1}{1+2018+2016\times2018}\)

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

\(=\frac{2017\times2018+1}{1+2018\times2017}\)

\(=1\)

\(\frac{2017.2018+1}{2019+2016.2018}\)

\(=\frac{2017.2018+1}{1+2018+2016.2018}\)

\(=\frac{2017.(2018+1)}{(1+2018).\left(2016+1\right)}\)

\(=\frac{2017.2019}{2019.2017}\)

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