TD
2
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
LM
31 tháng 10 2021
Bài nào khó lắm thì mới hỏi thôi chứ bài này dễ mà bạn tự vận động não đi
31 tháng 10 2021
\(\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{11\cdot12}=x\)
\(\Leftrightarrow x=\dfrac{1}{3}-\dfrac{1}{12}=\dfrac{4}{12}-\dfrac{1}{12}=\dfrac{1}{4}\)
T
5 tháng 8 2018
\(\frac{1}{5×6}+\frac{1}{6×7}+\frac{1}{7×8}+...+\frac{1}{24×25}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{24}-\frac{1}{25}\)
\(=\frac{1}{5}-\frac{1}{25}\)
\(=\frac{5}{25}-\frac{1}{25}\)
\(=\frac{4}{25}\)
VD
5 tháng 8 2018
\(\frac{1}{5\times6}+\frac{1}{6\times7}+\frac{1}{7\times8}+...+\frac{1}{24\times25}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{24}-\frac{1}{25}\)
\(=\frac{1}{5}-\frac{1}{25}\)
\(=\frac{4}{25}\)\(\left(=0,16\right)\)
ND
4
VN
Võ Ngọc Phương
CTVHS
VIP
30 tháng 8 2023
\(\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+...+\dfrac{1}{20\times21}=\dfrac{x}{14}\)
\(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{20}-\dfrac{1}{21}=\dfrac{x}{14}\)
\(\dfrac{1}{3}-\dfrac{1}{21}=\dfrac{x}{14}\)
\(\dfrac{7}{21}-\dfrac{1}{21}=\dfrac{x}{14}\)
\(\dfrac{6}{21}=\dfrac{x}{14}\)
\(\Rightarrow x.21=6.14\)
\(x.21=84\)
\(x=84:21\)
\(x=4\)
Vậy x = 4
30 tháng 8 2023
1/3x4 + 1/4x5 + 1/5x6 + .. + 1/20x21 = x/14
1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + .. + 1/20 - 1/21 = x/14
1/3 - 1/21 = x/14
7/21 - 1/21 = x/14
6/21 = x/14
x . 21 = 6 x 14
x x 21 = 84
x = 84 : 21
x = 4
WJ
1
HT
18 tháng 7 2015
\(\frac{1}{4.5}+\frac{1}{5.6}+.....+\frac{1}{99.100}\)
=\(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+.....+\frac{1}{99}-\frac{1}{100}\)
=\(\frac{1}{4}-\frac{1}{100}\)
=\(\frac{6}{25}\)
G
2
23 tháng 5 2023
1/3x4 + 1/4x5 + 1/5x6 +...+ 1/97x 98 + 1/98x99 + x =1
=> 1/3-1/4+1/4-1/5+1/5-1/6+....+1/97-1/98 + 1/98-1/99 +x = 1
=> 1/3 - 1/99 +x=1
=> 32/99+x=1
=> x= 1-32/99
=> x = 67/99
NQ
1
KN
10 tháng 1 2019
\(\text{Đặt }A=\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(\Leftrightarrow A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Leftrightarrow A=\frac{1}{2}-\frac{1}{100}\)
\(\Leftrightarrow A=\frac{49}{100}\)
10 tháng 1 2019
1/2x3+1/3x4+....+1/99x100
=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+....+1/99-1/100
=1-1/100
=99/100
NT
5
7 tháng 4 2022
= 1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+...+1/19-1/20
=1/2-1/20
=10/20-1/20
=9/20
7 tháng 4 2022
\(\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+...+\dfrac{1}{19\times20}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{19}-\dfrac{1}{20}\)
\(=\dfrac{1}{2}-\dfrac{1}{20}\)
\(=\dfrac{9}{20}\)
CH
1
BD
30 tháng 5 2023
\(\left(x+\dfrac{1}{2\times3}\right)+\left(x+\dfrac{1}{3\times4}\right)+\left(x+\dfrac{1}{4\times5}\right)+\left(x+\dfrac{1}{5\times6}\right)=\dfrac{25}{3}\)
\(x+\dfrac{1}{2\times3}+x+\dfrac{1}{3\times4}+x+\dfrac{1}{4\times5}+x+\dfrac{1}{5\times6}=\dfrac{25}{3}\)
\(x\times4+\left(\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}\right)=\dfrac{25}{3}\)
\(x\times4+\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}\right)=\dfrac{25}{3}\)
\(x\times4+\left(\dfrac{1}{2}-\dfrac{1}{6}\right)=\dfrac{25}{3}\)
\(x\times4+\dfrac{4}{12}=\dfrac{25}{3}\)
\(x\times4=\dfrac{25}{3}-\dfrac{4}{12}\)
\(x\times4=\dfrac{25}{3}-\dfrac{1}{3}\)
\(x\times4=\dfrac{24}{3}\)
\(x\times4=8\)
\(x=8\div4\)
\(x=2\)
:))
A = 1/3×5 + 1/5×7 + 1/7×9 + 1/9×11 + ... + 1/29×31
2A = 2/3×5 + 2/5×7 + 2/7×9 + 2/9×11 + ... + 2/29×31
2A = 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11 + .... + 1/29 - 1/31
2A = 1/3 - 1/31
2A = 31/93 - 3/93 = 28/93
A = 28/93 : 2
A = 28/93 × 1/2 = 14/93
B = 2/1×4 + 2/4×7 + 2/7×10 + ... + 2/31×34
3/2B = 3/1×4 + 3/4×7 + 3/7×10 + ... + 3/31×34
3/2B = 1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + ... + 1/31 - 1/34
3/2B = 1 -1/34 = 33/34
B = 33/34 : 3/2
B = 33/34 × 2/3 = 11/17
\(A=\frac{1}{3\times5}+\frac{1}{5\times7}+\frac{1}{7\times9}+...+\frac{1}{29\times31}\)
\(=\frac{1}{2}\times\left(\frac{1}{3}-\frac{1}{5}\right)+\frac{1}{2}\times\left(\frac{1}{5}-\frac{1}{7}\right)+...+\frac{1}{2}\times\left(\frac{1}{29}-\frac{1}{31}\right)\)
\(=\frac{1}{2}\times\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{29}-\frac{1}{31}\right)\)
\(=\frac{1}{2}\times\left(\frac{1}{3}-\frac{1}{31}\right)\)
\(=\frac{1}{2}\times\frac{28}{93}\)
\(=\frac{14}{93}\)
\(B=\frac{2}{1\times4}+\frac{2}{4\times7}+...+\frac{2}{31\times34}\)
\(=\frac{1}{3}\times\left(\frac{2}{1}-\frac{2}{4}\right)+\frac{1}{3}\times\left(\frac{2}{4}-\frac{2}{7}\right)+...+\frac{1}{3}\times\left(\frac{2}{31}-\frac{2}{34}\right)\)
\(=\frac{1}{3}\times\left(2-\frac{2}{4}+\frac{2}{4}-\frac{2}{7}+...+\frac{2}{31}-\frac{2}{34}\right)\)
\(=\frac{1}{3}\times\left(2-\frac{2}{34}\right)\)
\(=\frac{1}{3}\times\frac{33}{17}\)
\(=\frac{11}{17}\)