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b,
\(B=\frac{2018+2019}{2019+2020}=\frac{2018}{2019+2020}+\frac{2019}{2019+2020}\)
Ta thấy :
\(\frac{2018}{2019}>\frac{2018}{2019+2020}\)
\(\frac{2019}{2020}>\frac{2019}{2019+2020}\)
Từ đó , suy ra :
\(\frac{2018}{2019}+\frac{2019}{2020}>\frac{2018+2019}{2019+2020}\)
Vậy...
#Louis
Ta có :
\(\frac{205}{321}< \frac{205}{315}\)
\(\frac{214}{315}>\frac{205}{315}\)
\(\Leftrightarrow\frac{205}{321}< \frac{214}{315}\)
Ta có :
\(N=\frac{2018+2019+2020}{2019+2020+2021}\)
\(=\frac{2018}{2019+2020+2021}+\frac{2019}{2019+2020+2021}+\frac{2020}{2019+2020+2021}\)
Mà \(\frac{2018}{2019}>\frac{2018}{2019+2020+2021}\)
\(\frac{2019}{2020}>\frac{2019}{2019+2020+2021}\)
\(\frac{2020}{2021}>\frac{2020}{2019+2020+2021}\)
\(\Leftrightarrow M>N\)
Trả lời:
Ta có:
\(\frac{2018}{2019}>\frac{2018}{2019+2020+2021}\)
\(\frac{2019}{2020}>\frac{2019}{2019+2020+2021}\)
\(\frac{2020}{2021}>\frac{2020}{2019+2020+2021}\)
\(\Rightarrow\frac{2018}{2019}+\frac{2019}{2020}+\frac{2020}{2021}>\frac{2018+2019+2020}{2019+2020+2021}\)
hay \(M>N\)
Vậy \(M>N\)
Cho a,b,c dương thỏa mãn : a+b+c=2019
Tính A = \(\frac{a}{2019-c}+\frac{b}{2019-a}+\frac{c}{2019-b}\)
Thiếu dữ kiện, nếu chỉ cho vậy thì không tính đc gt cụ thể của A
+ Làm theo đề là tìm Min của A nhé!
\(A=\frac{a}{2019-c}+\frac{b}{2019-a}+\frac{c}{2019-b}=\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}.\)
\(A+3=\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}=\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\)\(\ge\left(a+b+c\right)\frac{9}{2\left(a+b+c\right)}=\frac{9}{2}\)(BĐT Bunhia)
Dấu "=" xra khi a=b=c=2019/3
\(4S=1+\frac{2}{4}+\frac{3}{4^2}+...+\frac{2019}{4^{2018}}.\)
\(4S-S=3S=1+\frac{2}{4}+\frac{3}{4^2}+...+\frac{2019}{4^{2018}}-\frac{1}{4}-\frac{2}{4^2}-...-\frac{2018}{4^{2018}}-\frac{2019}{4^{2019}}=1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{2018}}-\frac{2019}{4^{2019}}\)
\(3S< A=1+\frac{1}{4}+...+\frac{1}{4^{2018}}\)\(\Rightarrow3A=4A-A=4-\frac{1}{4^{2018}}< 4\)(sau khi rút gọn)
\(\Rightarrow3.3S< 4\Rightarrow9S< 4\)
\(\Rightarrow S< \frac{4}{9}< \frac{1}{2}\)
Ta có : \(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2019}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2018}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}+\frac{1}{2019}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2018}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1009}\right)\)
\(=\frac{1}{1010}+\frac{1}{1011}+\frac{1}{1012}+...+\frac{1}{2019}=B\)
\(\Rightarrow A-B-1=-1\)
\(\Rightarrow\left(A-B-1\right)^{2019}=-1\)
\(\frac{2019}{210}+\frac{2019}{280}+\frac{2019}{360}+\frac{2019}{450}+\frac{2019}{550}\)
\(=\frac{673}{70}+\frac{2019}{280}+\frac{673}{120}+\frac{673}{150}+\frac{2019}{550}\)
\(=\left[\frac{673}{70}+\frac{2019}{280}\right]+\frac{673}{120}+\frac{673}{150}+\frac{2019}{550}\)
\(=\left[\frac{2692}{280}+\frac{2019}{280}\right]+\frac{673}{120}+\frac{673}{150}+\frac{2019}{550}\)
\(=\frac{673}{40}+\frac{673}{120}+\frac{673}{150}+\frac{2019}{550}\)
\(=\left[\frac{673}{40}+\frac{673}{120}\right]+\frac{673}{150}+\frac{2019}{550}\)
\(=\left[\frac{2019}{120}+\frac{673}{120}\right]+\frac{673}{150}+\frac{2019}{550}\)
\(=\frac{673}{30}+\frac{673}{150}+\frac{2019}{550}\)
\(=\left[\frac{673}{30}+\frac{673}{150}\right]+\frac{2019}{550}\)
\(=\frac{673}{25}+\frac{2019}{550}=\frac{14806}{550}+\frac{2019}{550}=\frac{16825}{550}=\frac{673}{22}\)
P/S : Các a chị check dùm em ạ