So Sánh
A=\(\frac{11}{2017}+\frac{4}{2019}và\)B=\(\frac{10}{2017}+\frac{10}{2019}\)
K
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7 tháng 8 2018
\(a)\) Ta có :
\(VP=\frac{2018}{1}+\frac{2017}{2}+\frac{2016}{3}+...+\frac{2}{2017}+\frac{1}{2018}\)
\(VP=\left(\frac{2018}{1}-1-...-1\right)+\left(\frac{2017}{2}+1\right)+\left(\frac{2016}{3}+1\right)+...+\left(\frac{2}{2017}+1\right)+\left(\frac{1}{2018}+1\right)\)
\(VP=1+\frac{2019}{2}+\frac{2019}{3}+...+\frac{2019}{2017}+\frac{2019}{2018}\)
\(VP=2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}\right)\)
Lại có :
\(VT=\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\right).x\)
\(\Rightarrow\)\(x=2019\)
Vậy \(x=2019\)
Chúc bạn học tốt ~
23 tháng 6 2017
1. Bài giải:
Đặt \(A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{1000}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{1002}\)
\(\Rightarrow\frac{1}{2}A=A-\frac{1}{2}A=\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1000}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{1002}\right)\)
\(\Rightarrow\frac{1}{2}A=1-\frac{1}{1002}=\frac{1001}{1002}\Rightarrow A=\frac{2002}{1002}=\frac{1001}{501}\)
Vậy \(A=\frac{1001}{501}\)
T
17 tháng 2 2019
Ta có: \(B=\frac{10^2\left(10^{2017}+1\right)}{10^2\left(10^{2016}+1\right)}=\frac{10^{2019}+1+99}{10^{2018}+1+99}\)
Do phân số \(A=\frac{10^{2019}+1}{10^{2018}+1}>1\).Áp dụng BĐT \(\frac{a}{b}>1\Rightarrow\frac{a}{b}>\frac{a+m}{b+m}\left(m>0\right)\).
Ta có: \(A=\frac{10^{2019}+1}{10^{2018}+1}>\frac{10^{2019}+1+99}{10^{2018}+1+99}=B\)
Vậy \(A>B\)
27 tháng 6 2019
\(A=\frac{2016}{2017}+\frac{2017}{2018}+\frac{2018}{2019}\)
\(\Rightarrow A=(1-\frac{1}{2017})+(1-\frac{1}{2018})+(1-\frac{1}{2019})\)
\(\Rightarrow A=3-\left(\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}\right)\)
\(\left(\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}\right)\)<\(\frac{3}{2017}\)<\(1\)
\(\Rightarrow A\)>\(3-1=2\)
\(B=\frac{2016+2017+2018}{2017+2018+2019}\)
\(\Rightarrow B=1-\frac{3}{6054}\)
\(\Rightarrow B=1-\frac{1}{2018}\)
\(B\)<\(1\);\(A\)>\(2\)
\(\Rightarrow A\)>\(B\)
12 tháng 8 2019
\(Ta\)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)
KV
18 tháng 3 2019
ta có :
\(A=\frac{10^{2019}+1}{10^{2018}+1}=\frac{10^{2018}.10+1}{10^{2018}+1}=\frac{10}{10^{2018}+1}\)
\(B=\frac{10^{2018}+1}{10^{2017}+1}=\frac{10^{2017}.10+1}{10^{2017}+1}=\frac{10}{10^{2017}+1}\)
Do \(10^{2017}+1< 10^{2018}+1\Rightarrow\frac{10}{10^{2017}+1}>\frac{10}{10^{2018}+1}\)
\(\Rightarrow A< B\)
\(A=\frac{2016^{2016}+1}{2016^{2017}+1}\Rightarrow2016A=\frac{2016^{2017}+2016}{2016^{2017}+1}=1+\frac{2015}{2016^{2017}+1}\)
\(B=\frac{2016^{2017}-3}{2016^{2018}-3}\Rightarrow2016B=\frac{2016^{2018}-6048}{2016^{2018}-3}=1+\frac{-6045}{2016^{2018}-3}\)
Vì \(\frac{2015}{2016^{2017}+1}>0;\frac{-6045}{2016^{2018}-3}< 0\)
Nên: A>B