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\(M=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{37.38.39}=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+..+\frac{1}{37.38}-\frac{1}{38.39}\)
\(=\frac{1}{1.2}-\frac{1}{38.39}=\frac{1}{1}-\frac{1}{2}-\frac{1}{38}+\frac{1}{39}=\frac{370}{741}\)
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}=\frac{1}{k}.\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)=\frac{1}{k}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)=\frac{1}{k}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(\Leftrightarrow\frac{1}{2}=\frac{1}{k}\Rightarrow k=2\)
A =
A = \(1-\frac{1}{2018}\)
A = \(\frac{2017}{2018}\)
Có :
2.B = \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2015.2017}\)
2.B = \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2015}-\frac{1}{2017}\)
2.B = \(1-\frac{1}{2017}\)
2.B = \(\frac{2016}{2017}\)
B = \(\frac{2016}{2017}:2=\frac{1008}{2017}\)
Có :
3.C = \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{2017.2020}\)
3.C = \(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{2017}-\frac{1}{2020}\)
3.C = \(\frac{1}{1}-\frac{1}{2020}=\frac{2019}{2020}\)
C = \(\frac{2019}{2020}:3=\frac{673}{2020}\)
đặt 22018 = a ; 32019 = b ; 52020 = c
Ta có : \(A=\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{a+c}>\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=1\)
\(B=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\)
\(2B=\frac{2}{1.2}+\frac{2}{3.4}+...+\frac{2}{2019.2020}\)
\(< 1+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018.2019}+\frac{1}{2019.2020}\)
\(2B< 1+\frac{3-2}{2.3}+\frac{4-3}{3.4}+....+\frac{2019-2018}{2018.2019}+\frac{2020-2019}{2019.2020}\)
\(2B< 1+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}=1+\frac{1}{2}-\frac{1}{2020}< 1+\frac{1}{2}\)
\(B< \frac{3}{4}\)
\(\Rightarrow A>1>\frac{3}{4}>B\)
Mình chỉ biết cách tính B thôi, đây nhé:
B= \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{2019.2020}\)
B=\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{2019}-\frac{1}{2020}\)
\(B=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2019}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2020}\right)\)
\(B=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2019}+\frac{1}{2020}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2020}\right)\)
\(B=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2019}+\frac{1}{2020}\right)-2\frac{1}{2}\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1010}\right)\)
\(B=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2019}+\frac{1}{2020}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1010}\right)\)
\(B=\frac{1}{1011}+\frac{1}{1012}+....+\frac{1}{2019}+\frac{1}{2020}\)
\(A=\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(\frac{A}{7}=\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(\frac{A}{7}=\frac{7-2}{2.7}+\frac{11-7}{7.11}+\frac{14-11}{11.4}+\frac{15-14}{14.15}+\frac{28-15}{15.28}\)
\(\frac{A}{7}=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}=\frac{1}{2}-\frac{1}{28}=\frac{13}{28}\)
\(A=7.\frac{13}{28}\)
\(A=\frac{13}{4}\)
Trước hết ta thực hiện biểu thức trong ngoặc:
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{8.9.10}\)
\(=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{8.9.10}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{9.10}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{90}\right)\)
\(=\frac{1}{2}.\frac{22}{45}\) \(=\frac{11}{45}\)
\(\Rightarrow\frac{11}{45}\) \(.x=\frac{22}{45}\)
\(\Rightarrow x=\frac{22}{45}:\frac{11}{45}\)
\(\Rightarrow x=2\)
a) \(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2019\cdot2020}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\)
\(A=\frac{1}{1}-\frac{1}{2020}\)
\(A=\frac{2020}{2020}-\frac{1}{2020}=\frac{2019}{2020}\)
b) \(\frac{1}{m-1}-\frac{1}{m}=\frac{m}{m\left(m-1\right)}-\frac{m-1}{m\left(m-1\right)}=\frac{m-\left(m-1\right)}{m\left(m-1\right)}=\frac{m-m+1}{m\left(m-1\right)}=\frac{1}{m\left(m-1\right)}=\frac{1}{m^2-m}\)
Dễ dàng nhận thấy \(m^2>m^2-m\)
\(\Rightarrow\frac{1}{m^2}< \frac{1}{m^2-m}\)hay \(\frac{1}{m^2}< \frac{1}{m-1}-\frac{1}{m}\)( với m thuộc N , m > 1 ) ( đpcm )
a) \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\)
\(=1-\frac{1}{2020}=\frac{2019}{2020}\)
Vậy \(A=\frac{2019}{2020}\).
b) \(\frac{1}{m-1}-\frac{1}{m}=\frac{m}{m\left(m-1\right)}-\frac{m-1}{m\left(m-1\right)}=\frac{m-\left(m-1\right)}{m^2-m}=\frac{1}{m^2-m}\)
Mà \(m\inℕ;m>1\) nên \(m^2>m^2-m\)
\(\Rightarrow\frac{1}{m}< \frac{1}{m-1}-\frac{1}{m}\) (đpcm)