A=1/1.2+1/2.3+1/3.4+1/4.5+....+1/2018.2019
Cho A=1/1.2+1/2.3+1/3.4+1/4.5+...+1/2018.2019+1/2019.2020 thì A có giá trị là ?
Giúp mình với ạ mình đang cần gấp í:)
\(\text{#}HaimeeOkk\)
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{2018.2019}+\dfrac{1}{2019.2020}\)
\(A=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{2018}-\dfrac{1}{2019}+\dfrac{1}{2019}-\dfrac{1}{2020}\)
\(A=1-\left(\dfrac{1}{2}-\dfrac{1}{2}\right)-\left(\dfrac{1}{3}-\dfrac{1}{3}\right)-\left(\dfrac{1}{4}-\dfrac{1}{4}\right)-...-\left(\dfrac{1}{2019}-\dfrac{1}{2019}\right)-\dfrac{1}{2020}\)
\(A=1-0-0-0-...-0-\dfrac{1}{2020}\)
\(A=1-\dfrac{1}{2020}\)
\(A=\dfrac{2019}{2020}\)
Vậy \(A=\dfrac{2019}{2020}\)
Tính tổng sau một cách hợp lí :
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2017.2018}+\frac{1}{2018.2019}\)
Làm được t tick
gọi biểu thức trên là A A=1/1 -1/2+1/3-1/4+...+1/2017-12018+1/2018-1/2019 A=1/1-1/2019 A=2018/2019
1/1.2+1/2.3+1/3.4+1/4.5+...+1/2017.2018+1/2018.2019
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2018}-\frac{1}{2019}\)
\(=1-\frac{1}{2019}\)
\(=\frac{2019}{2019}-\frac{1}{2019}\)
\(=\frac{2018}{2019}\)
cái ĐỒ ĐÁNG GHÉT ◥ὦɧ◤ŤŔầŃ VăŃ ĤùŃĞ™ kia t định trả lời sao m dám....
A=1/1.2+1/2.3+1/3.4+...+1/2018.2019
Chứng tỏ A<1
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018+2019}\)
=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2018}-\frac{1}{2019}\)
=\(1-\frac{1}{2019}< 1\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018.2019}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2018}-\frac{1}{2019}\)
\(A=\frac{1}{1}-\frac{1}{2019}< 1\)
Vậy \(A< 1\)
S = 1/1.2 + 1/2.3 + 1/3.4 +..........1/2017.2018 + 1/2018.2019
can gap
\(S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}+\frac{1}{2018.2019}\)
\(S=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2018}-\frac{1}{2019}\)
( gạch bỏ các phân số giống nhau)
\(S=1-\frac{1}{2019}\)
\(S=\frac{2018}{2019}\)
CHÚC BN HỌC TỐT!!!!
S=1/1.2+1/2.3+1/3.4+............1/2017.2018+1/2018.2019
S=1/2.(1+1/3.2+1/3.2+.............1/2017.1009+1/1009.2019)
S=1/4.(2+2/3.2+2/3.2+..............2/2017.1009+2/1009.2019)
S=1/4.(1-1/2+1/2-1/3+1/3+..........+1/1009-1/1009+1/2019)
S=1/4.(1-1/2019)
S=1/4.2018/2019=1009/4038
S=1-1/2+1/2-1/3+1/3-1/4+.........+1/2017-1/2018+1/2018-1/2019
S=1-1/2019
S=2019/2019-1/2019
S=2018/2019
Đề bài:
So sánh:
A=1/1.2+1/2.3+1/3.4+1/4.5+...+1/2018.2019 với 1
Ai nhanh và đúng, mk tk cho!!!
Và m.n tk lại mk nhé
\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{2018\cdot2019}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2018}-\frac{1}{2019}\)
\(A=1-\frac{1}{2019}=\frac{2018}{2019}\)
Mà \(\frac{2018}{2019}< \frac{2019}{2019}=1\)
\(\Rightarrow A< 1\)
#)Giải :
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2018}-\frac{1}{2019}\)
\(A=1-\frac{1}{2019}\)
\(A=\frac{2018}{2019}\)
Vì \(\frac{2018}{2019}< 1\Rightarrow A< 1\)
#~Will~be~Pens~#
\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2018.2019}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2018}-\frac{1}{2019}\)
\(=1-\frac{1}{2019}< 1\)
\(\Leftrightarrow A< 1\)
tìm x biết :(1.2+2.3+3.4+...+2017.2018)/(2018.2019.x)=1/(1+2)+1/(1+2+3)+....+1/(1+2+....+2018)
Cho A=1/1.2 + 1/2.3 + + 1/ 3.4+...+1/49.50 ; B = 1.2+2.3+3.4+4.5+5.6+...+49.50
Tính 50 mủ 2 A – B/17
A = 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(A=1-\frac{1}{6}\)
\(A=\frac{5}{6}\)
_Chúc bạn học tốt_
Ta có : \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}\)
\(=1-\frac{1}{5}=\frac{4}{5}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}_{ }-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}\)
\(A=1-\frac{1}{6}\)
\(A=\frac{5}{6}\)
a=1/1.2=1/2.3+1/3.4+1/4.5+....+1/49.5
bạn ơi đánh có dấu đi mình ko hiểu ý bạn là gì