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Ta có :
\(\frac{2017\times2018+1}{2019+2016\times2018}\)
\(=\frac{2017\times2018+1}{1+2018+2016\times2018}\)
\(=\frac{2017\times2018+1}{1+2018\times\left(2016+1\right)}\)
\(=\frac{2017\times2018+1}{1+2018\times2017}\)
\(=1\)
\(\frac{2017.2018+1}{2019+2016.2018}\)
\(=\frac{2017.2018+1}{1+2018+2016.2018}\)
\(=\frac{2017.(2018+1)}{(1+2018).\left(2016+1\right)}\)
\(=\frac{2017.2019}{2019.2017}\)
\(=\frac{1}{1}=1\)
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!
\(\Rightarrow\text{❤️✔✨♕✨✔️❤ }\Leftarrow\)
\(\text{Ta có :}\)
\(A=\frac{2017\cdot2018-1}{2017\cdot2018-2}=\frac{4070305}{4070304}=1\frac{1}{4070304}\)
\(B=\frac{2017}{2018}\)
\(\text{Vì : }1\frac{1}{4070304}>1\text{ mà }\frac{2017}{2018}< 1\text{ nên }1\frac{1}{4070304}>\frac{2017}{2018}\)
\(\Rightarrow A>B\)
\(A=\frac{2019}{2}+\frac{2019}{6}+\frac{2019}{12}+....+\frac{2019}{2018.2019}\)
\(=\frac{2019}{1}.\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{2018.2019}\right)\)
\(=\frac{2019}{1}.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018.2019}\right)\)
\(=\frac{2019}{1}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{4}+....+\frac{1}{2018}-\frac{1}{2019}\right)\)
\(=\frac{2019}{1}.\left(1-\frac{1}{2019}\right)\)
\(=\frac{2019}{1}.\frac{2018}{2019}\)
\(=2018\)
\(A=\frac{2019}{2}+\frac{2019}{6}+\frac{2019}{12}+\frac{2019}{20}+\frac{2019}{30}+\frac{2019}{2018.2019}\)
\(A=\frac{2019}{1.2}+\frac{2019}{2.3}+\frac{2019}{3.4}+\frac{2019}{4.5}+\frac{2019}{5.6}+...+\frac{2019}{2018.2019}\)
\(A=2019.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018.2019}\right)\)
\(A=2019.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2018}-\frac{1}{2019}\right)\)
\(A=2019.\left(1-\frac{1}{2019}\right)\)\(=2019.\frac{2018}{2019}=2018\)
Vậy A = 2018
-Dấu " . " là dấu nhân.
\(Ta\)có :\(a\)=\(\frac{2017\cdot2018-1}{2017.2018}\)=\(\frac{2017.2018}{2017.2018}\)-\(\frac{1}{2017.2018}\)=1-\(\frac{1}{2017.2018}\)
\(b\)=\(\frac{2019.2020-1}{2019.2020}\)=\(\frac{2019.2020}{2019.2020}\)-\(\frac{1}{2019.2020}\)=1-\(\frac{1}{2019.2020}\)
Vì \(\frac{1}{2018.2019}\)> \(\frac{1}{2019.2020}\)nên \(a\)< \(b\)(sử dụng phần bù)
so sánh a và b biết a=2017×2018−12017×20182017×20182017×2018−1và b =2019×2020−12019×20202019×20202019×2020−1
\(1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\)
\(=1+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\right)\)
\(=1+\left(1-\frac{1}{2018}\right)\)
\(=1+\left(\frac{2018}{2018}-\frac{1}{2018}\right)\)
\(=1+\left(\frac{2017}{2018}\right)\)
\(=\frac{2018}{2018}+\frac{2017}{2018}=\frac{4035}{2018}\)
\(1+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}...+\frac{1}{2017\cdot2018}\)
\(=1+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}...+\frac{1}{2017}-\frac{1}{2018}\right)\)
\(=1+\left(1-\frac{1}{2018}\right)\)
\(=1+\frac{2017}{2018}\)
\(=1+\frac{2017}{2018}\)
\(=\frac{4035}{2018}\)
các bạn hãy trả lời nhanh cho mình với nhé
a)
2018x2019- 2017 = 2017 = 1 b) 2018 x 2018 =2018x 2018 = 20182017 x
2018+20192017 (2017 + 1) x 20192018x 2019 2019chúc bạn hok tốt