Chứng minh \(\frac{A}{B}\) là số nguyên với :
\(A=\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+...+\frac{1}{2015\cdot2016}+\frac{1}{2017\cdot2018}\)
\(B=\frac{1}{1010\cdot2018}+\frac{1}{1011\cdot2017}+...+\frac{1}{2018\cdot1010}\)
Tính A=\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+..........+\frac{1}{2017\cdot2018}\)
B=\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+..........+\frac{1}{2015\cdot2017}\)
C=\(\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+..........+\frac{1}{2017\cdot2020}\)
Ai làm nhanh nhất mình sẽ Tick cho nha.
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 = \(\frac{2019}{2020}:3=\frac{673}{2020}\)
a=1/1-1/2+1/2-1/3+...+1/2017-1/2018
=1/1-1/2018
=kq
may bai duoi lam tuong tu nha
mình chưa điền kết quả ban tu dien nha
\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2017\cdot2018}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(A=\frac{1}{1}-\frac{1}{2018}=\frac{2017}{2018}\)
\(B=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{2015\cdot2017}\)
\(B=\frac{1}{2}\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{2015\cdot2017}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{2017}\right)\)
\(B=\frac{1}{2}\cdot\frac{2016}{2017}\)
\(B=\frac{1008}{2017}\)
\(C=\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+...+\frac{1}{2017\cdot2020}\)
\(C=\frac{1}{3}\left(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{2017\cdot2020}\right)\)
\(C=\frac{1}{3}\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{2017}-\frac{1}{2020}\right)\)
\(C=\frac{1}{3}\left(\frac{1}{1}-\frac{1}{2020}\right)\)
\(C=\frac{1}{3}\cdot\frac{2019}{2020}\)
\(C=\frac{673}{2020}\)
1.
a) \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{2017\cdot1018}\)
b) \(\frac{2}{1\cdot2}+\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+....+\frac{2}{2017\cdot2018}+\frac{2}{2018\cdot2019}\)
c) \(\frac{4}{1\cdot2}+\frac{4}{2\cdot3}+\frac{4}{3\cdot4}+......+\frac{4}{1999\cdot2000}\)
********Lưu ý :
1) Các dấu chấm ở phân số là dấu nhân ạ..!!!
2) Trình bày rõ ràng giúp mk với ạ..!!!
a) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.......+\frac{1}{2017.2018}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-..........-\frac{1}{2018}\)
\(=1-\frac{1}{2018}\)
\(=\frac{2018}{2018}-\frac{1}{2018}=\frac{2017}{2018}\)
b) \(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+..........+\frac{2}{2017.2018}+\frac{2}{2018.2019}\)
\(=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.........+\frac{1}{2017.2018}+\frac{1}{2018.2019}\right)\)
\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-.........-\frac{1}{2018}+\frac{1}{2018}-\frac{1}{2019}\right)\)
\(=2\left(1-\frac{1}{2019}\right)\)
\(=2\left(\frac{2019}{2019}-\frac{1}{2019}\right)\)
\(=2.\frac{2018}{2019}\)
\(=\frac{4036}{2019}\)
Phần c tương tự nha
a) \(\frac{1}{1.2}\) + \(\frac{1}{2.3}\) + .......+ \(\frac{1}{2017.2018}\)
= 1 - \(\frac{1}{2}\) + \(\frac{1}{2}\) - \(\frac{1}{3}\) + .......+ \(\frac{1}{2017}\) - \(\frac{1}{2018}\)
= 1 - \(\frac{1}{2018}\) = \(\frac{2017}{2018}\)
câu a) mik sửa đề một tí ko biết có đúng ko
câu b , c tương tự nhưng cần lấy tử ra chung
a)\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{2017\times2018}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(=1-\frac{1}{2018}\)
\(=\frac{2017}{2018}\)
b)nhóm 2 ra ngoài rồi làm như câu a
c)nhóm 4 ra rồi làm như câu a
a)\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{2017\cdot2018}\) b)\(\left[x\cdot\frac{5}{3}-1\right]:9=3\frac{1}{2}:2,25\)
dấu chấm là dấu nhân nha :3
\(a,\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2017\cdot2018}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(=1-\frac{1}{2018}\)
\(=\frac{2017}{2018}.\)
\(b,\left[x\cdot\frac{5}{3}-1\right]:9=3\frac{1}{2}:2,25\)
\(\Leftrightarrow\left[x\cdot\frac{5}{3}-1\right]:9=\frac{7}{2}:\frac{9}{4}\)
\(\Leftrightarrow\left[x\cdot\frac{5}{3}-1\right]:9=\frac{7}{2}\cdot\frac{4}{9}\)
\(\Leftrightarrow\left[x\cdot\frac{5}{3}-1\right]:9=\frac{14}{9}\)
\(\Leftrightarrow x\cdot\frac{5}{3}-1=\frac{14}{9}\cdot9\)
\(\Leftrightarrow x\cdot\frac{5}{3}-1=14\)
\(\Leftrightarrow x\cdot\frac{5}{3}=14+1\)
\(\Leftrightarrow x\cdot\frac{5}{3}=15\)
\(\Leftrightarrow x=15:\frac{5}{3}\)
\(\Leftrightarrow x=15\cdot\frac{3}{5}\)
\(\Leftrightarrow x=9.\)
a)\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2017.2018}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(=\frac{1}{1}-\frac{1}{2018}\)
\(=\frac{2017}{2018}\)
b)\(\left[x.\frac{5}{3}-1\right]:9=3\frac{1}{2}:2,25\)
\(\Leftrightarrow\left[x.\frac{5}{3}-1\right]:9=3\frac{1}{2}:\frac{9}{4}=1\frac{5}{9}\)
\(\Rightarrow x.\frac{5}{3}-1=1\frac{5}{9}.9=14\)
\(\Rightarrow x.\frac{5}{3}=14+1=15\)
\(\Rightarrow x=15:\frac{5}{3}=9\)
a) \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2017\cdot2018}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(=\frac{1}{1}-\frac{1}{2018}\)
\(=\frac{2018}{2018}-\frac{1}{2018}\)
\(=\frac{2017}{2018}\)
b) \(\left(x\cdot\frac{5}{3}-1\right):9=3\frac{1}{2}:2,25\)
\(\left(x\cdot\frac{5}{3}-1\right):9=\frac{7}{2}:\frac{9}{4}\)
\(\left(x\cdot\frac{5}{3}-1\right):9=\frac{14}{9}\)
\(x\cdot\frac{5}{3}-1=\frac{14}{9}\cdot9\)
\(x\cdot\frac{5}{3}-1=14\)
\(x\cdot\frac{5}{3}=14+1\)
\(x\cdot\frac{5}{3}=15\)
\(x=15:\frac{5}{3}\)
\(x=9\)
Vậy \(x=9\)
Tính: 1-\(\frac{1}{1\cdot2}\)-\(\frac{1}{2\cdot3}\)-\(\frac{1}{3\cdot4}\)-...-\(\frac{1}{2017\cdot2018}\)
\(1-\frac{1}{1.2}-\frac{1}{2.3}-\frac{1}{3.4}-....-\frac{1}{2017.2018}\)
\(=1-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\right)\)
\(=1-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2017}-\frac{1}{2018}\right)\)
\(=1-\left(1-\frac{1}{2018}\right)\)
\(=1-1+\frac{1}{2018}=\frac{1}{2018}\)
cho hỏi : A=\(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{2015\cdot2016}\)
B=\(\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\)
tính B-A
giúp vs nha,cảm ơn nhìu!
A\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{2015}-\frac{1}{2016}\)
A=\(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2015}-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2016}\right)\)
A=\(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2015}+\frac{1}{2016}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2016}\right)\)
A=\(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2015}+\frac{1}{2016}-\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1008}\right)\)
A=\(\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2015}+\frac{1}{2016}\)
B-A=\(\left(\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\right)-\left(\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2015}+\frac{1}{2016}\right)\)
B-A=1/1008
cho hỏi: A=\(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{2015\cdot2016}\)
B=\(\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\)
Tính B-A
Ai giúp cảm ơn!
Trường THCS Lý Tự Trọng
Đề thi khảo sát chọn lớp dành cho học sinh thi vào lớp 6 môn toán
Bài 1 : Tính ( tính nhanh nếu có thể )
a) 1001 x 789 + 456 x 128 - 789 + 912 x 436
b) \(\left(2016\cdot2017+2017\cdot2018\right)\cdot\left(1+\frac{1}{2}:1\frac{1}{2}-1\frac{1}{3}\right)\)
c) \(5\frac{9}{10}:\frac{3}{2}-\left(2\frac{1}{3}\cdot4\frac{1}{2}-2\cdot2\frac{1}{3}\right):\frac{7}{4}\)
b)\(\left(2016.1017+2017.2018\right).\left(1+\frac{1}{2}:\frac{3}{2}-\frac{4}{3}\right)\)
\(\left(2016.2017+2017.2018\right)\left(1+\frac{1}{3}-\frac{4}{3}\right)\)
\(\left(2016.2017+2017.2018\right).\left(\frac{4}{3}-\frac{4}{3}\right)\)
\(\left(2016.2017+2017.2018\right).0\)
\(=0\)
a) \(1001.789+456.128.128-789+912.436\)
\(=\left(1001.789-789\right)+\left(456.2.64.128+912.436\right)\)
\(=789.1000+912.4\left(2048+109\right)\)
\(=789000+912.4.2157\)
\(=8657736\)
c)\(5\frac{9}{10}+\frac{3}{2}-\left(2\frac{1}{3}.4\frac{1}{2}-2.2\frac{1}{3}\right):\frac{7}{4}\)
\(=\frac{59}{10}+\frac{3}{2}-\left(\frac{7}{3}.\frac{9}{2}-2.\frac{7}{3}\right):\frac{7}{4}\)
\(=\frac{59}{10}+\frac{3}{2}-\left[\frac{7}{3}\left(\frac{9}{2}-2\right)\right]:\frac{7}{4}\)
\(=\frac{59}{10}+\frac{3}{2}-\left(\frac{7}{3}.\frac{5}{2}\right):\frac{7}{4}\)
\(=59+\frac{3}{2}-\frac{35}{6}.\frac{4}{7}\)
\(=\frac{59}{10}+\frac{3}{2}-\frac{10}{3}\)
\(=\frac{177+45-100}{30}=\frac{122}{30}=\frac{61}{15}\)
A=\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2017}\)
B=\(\frac{1}{1009}+\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2016}+\frac{1}{2017}\)
tính (A-B)^2017
1,So sánh A và B
\(A=\frac{2017\cdot2018-1}{2017\cdot2018}\)
\(B=\frac{2018\cdot2019-1}{2018\cdot2019}\)
\(A=\frac{2017.2018-1}{2017.2018}=1-\frac{1}{2017.2018}\)
\(B=\frac{2018.2019-1}{2018.2019}=1-\frac{1}{2018.2019}\)
Có \(\frac{1}{2017.2018}>\frac{1}{2018.2019}\)
\(\Rightarrow A< B\)
\(A=\frac{2017.2018-1}{2017.2018}=1-\frac{1}{2017.2018}\)
\(B=\frac{2018.2019-1}{2018.2019}=1-\frac{1}{2018.2019}\)
Do \(\frac{1}{2017.2018}>\frac{1}{2018.2019}\)nên \(1-\frac{1}{2017.2018}< 1-\frac{1}{2018.2019}\)
Vậy \(A< B\)