Tìm D biết:
D=(1/1x2)+(1/2x3)+(1/3x4)+...+(1/2018x2019)
lẹ lẹ tý coi
1/1x2 + 1/2x3 + 1/3x4 + .......+1/2018x2019
giải đầy đủ
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{2018.2019}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2018}-\frac{1}{2019}\)
\(=1-\frac{1}{2019}\)
\(=\frac{2018}{2019}\)
Dấu \(.\)là dấu nhân .
Ta có :
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018.2019}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2018}-\frac{1}{2019}\)
\(=1-\frac{1}{2019}\)
\(=\frac{2019}{2019}-\frac{1}{2019}\)
\(=\frac{2018}{2019}\)
~ Ủng hộ nhé
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018.2019}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2018}-\frac{1}{2019}\)
\(=1-\frac{1}{2019}\)
\(=\frac{2018}{2019}\)
. Tìm x biết :
a) 1/1x2 + 1/2x3 + 1/3x4 + ... + 1/x.(x+1) = 667/668
b) 1/1.4 + 1/4.7 + 1/7.10 + ... + 1/x.(x+3) = 667/2002
c) 1/3 + 1/6 + 1/10 + 1/15 + ... + 1/x.(2x+1) = 1999/2001
lẹ giùm mình T.T mình sắp đi học ròii nên ai nhanh mình tick
a) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x.\left(x+1\right)}=\frac{667}{668}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{667}{668}\)
\(1-\frac{1}{x+1}=\frac{667}{668}\)
\(\frac{1}{x+1}=1-\frac{667}{668}\)
\(\frac{1}{x+1}=\frac{1}{668}\)
\(\Rightarrow x+1=668\)
x = 667
a) 1/1x2 + 1/2x3 + 1/3x4 + ... + 1/x.(x+1) = 667/668
=>1/1-1/2+1/2-1/3+1/3-1/4+.......+1/x-1/x+1=667/668
=>1/1-1/x+1=667/668
=>1/x+1=1/1-667/668
=>1/x+1=1/668
=>x=667
b) \(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x.\left(x+3\right)}=\frac{667}{2002}\)
\(\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{667}{2002}\)
\(\frac{1}{3}.\left(1-\frac{1}{x+3}\right)=\frac{667}{2002}\)
\(1-\frac{1}{x+3}=\frac{2001}{2002}\)
\(\frac{1}{x+3}=\frac{1}{2002}\)
=> x + 3 = 2002
x = 1999
C=\(\dfrac{2}{1x2}\)+\(\dfrac{2}{2x3}\)+\(\dfrac{2}{3x4}\)+...+\(\dfrac{2}{2018x2019}\)+\(\dfrac{2}{2019x2020}\)
\(C=\dfrac{2}{1\times2}+\dfrac{2}{2\times3}+...+\dfrac{2}{2019\times2020}\)
\(=2\left(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+...+\dfrac{1}{2019\times2020}\right)\)
\(=2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2019}-\dfrac{1}{2020}\right)\)
\(=2\left(1-\dfrac{1}{2020}\right)=2.\dfrac{2019}{2020}=\dfrac{2019}{1010}\)
lớp 5 đây á
no no
đây ko phải lớp 5 mọi người nhỉ ?
1/1x2+1/2x3+1/3x4+1/24x25
1/1x2+ 1/2x3+1/3x4+1/24x25
\(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+....+\dfrac{1}{24\times25}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{24}-\dfrac{1}{25}\)
\(=1-\dfrac{1}{25}\)
\(=\dfrac{24}{25}\)
Câu 1: Thực hiện phép tính
A= -125 x 2^3 + 71 x 53 + 53 x (- 29) - 42 x 53
Câu 2: Tính giá trị biểu thức
A= 2019/1x2 + 2019/2x3 + 2019/3x4 +...........+ 2019/2018x2019
Tính giá trị biểu thức
A= 2019/1x2 + 2019/2x3 + 2019/3x4 +.............+ 2019/2018x2019
Ai nhanh m tick nha
\(\frac{2019}{1\times2}+\frac{2019}{2\times3}+\frac{2019}{3\times4}+...+\frac{2019}{2018\times2019}\)
\(=2019\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{2018\times2019}\right)\)
\(=2019\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2018}-\frac{1}{2019}\right)\)
\(=2019\left(1-\frac{1}{2019}\right)\)
\(=2019\left(\frac{2019}{2019}-\frac{1}{2019}\right)\)
\(=2019\times\frac{2018}{2019}\)\(=\frac{2019\times2018}{2019}=2018\)
Tính tổng
a) 1/2x3 + 1/3x4 + 1/4x5 + ... + 1/2018x2019
b) 1/2 + 1/4 + 1/8 + 1/16 + ... + 1/2048
(1/(1x2)/(2x3)/(3x4)):(1/(2x3)/(3x4)/(4x5)):...(1/(97*98)/(98*99)/(99*100))
(1/(1x2)/(2x3)/(3x4)):(1/(2x3)/(3x4)/(4x5)):...(1/(97*98)/(98*99)/(99*100
haizzz đáng tiếc tôi muốn ns là: ko bao f và đừng mong chờ OK