Ta có công thức tổng quát:

\(1+\frac{1}{\left(n-1\right)\left(n+1\right)}\)

\(=1+\frac{1}{n^2-1}\)

\(=\frac{n^2-1+1}{n^2-1}=\frac{n^2}{\left(n-1\right)\left(n+1\right)}\)

\(\left(1+\frac{1}{1\cdot3}\right)\left(1+\frac{1}{2\cdot4}\right)\cdot\ldots\cdot\left(1+\frac{1}{2019\cdot2021}\right)\)

\(=\left(1+\frac{1}{\left(2-1\right)\left(2+1\right)}\right)\left(1+\frac{1}{\left(3-1\right)\left(3+1\right)}\right)\cdot\ldots\cdot\left(1+\frac{1}{\left(2020-1\right)\left(2020+1\right)}\right)\)

\(=\frac{2^2-1+1}{\left(2-1\right)\left(2+1\right)}\cdot\frac{3^2-1+1}{\left(3-1\right)\left(3+1\right)}\cdot\ldots\cdot\frac{2020^2-1+1}{\left(2020-1\right)\left(2020+1\right)}\)

\(=\frac{2^2}{1\cdot3}\cdot\frac{3^2}{2\cdot4}\cdot\ldots\cdot\frac{2020^2}{2019\cdot2021}\)

\(=\frac{2\cdot3\cdot\ldots\cdot2020}{1\cdot2\cdot3\cdot\ldots\cdot2019}\cdot\frac{2\cdot3\cdot\ldots\cdot2020}{3\cdot4\cdot\ldots\cdot2021}=\frac{2020}{1}\cdot\frac{2}{2021}=\frac{4040}{2021}\)

A = (1+ 1/1.3).(1 + 1/2.4)...(1+1/2019.2021)

A = \(\frac{1.3+1}{1.3}\).\(\frac{2.4+1}{2.4}\)...\(\frac{2009.2021+1}{2009.2001}\)

A = \(\frac{4}{1.3}\).\(\frac{9}{2.4}\)...\(\frac{4080400}{2009.2021}\)

A = \(\frac{2.2}{1.3}\).\(\frac{3.3}{2.4}\)...\(\frac{2020.2020}{2009.2021}\)

A = \(\frac{2.3...2020}{1.2\ldots2009}\) . \(\frac{2.3.4\ldots2020}{3.4.\ldots2021}\)

A = \(\frac{2020.2}{1.2021}\)

A = \(\frac{4040}{2021}\)

0,75 - (2\(\frac13\) + 0,75) + 3^2.(- \(\frac19\))

= 0,75 - 2\(\frac13\) - 0,75 - 1

= (0,75 - 0,75) - (2\(\frac13\) + 1)

= 0 - (\(\frac73+1\))

= 0 - (\(\frac73+\frac33\))

= 0 - 10/3

= - 10/3

-3/5 : 7/5 - 3/5 : 7/5 + 2\(\frac35\)

= (-3/5 - 3/5) : 7/5 +

= -\(\frac65\) x \(\frac57\) + \(\frac{13}{5}\)

= - 6/7 + \(\frac{13}{5}\)

= - \(\frac{30}{35}\) + \(\frac{91}{35}\)

= 61/35

\(-\frac35:\frac75-\frac35:\frac75+2\frac35\)

\(=-\frac35\cdot\frac57-\frac35\cdot\frac57+2+\frac35\)

\(=-\frac37-\frac37+2+\frac35=-\frac67+\frac{13}{5}=\frac{-30+91}{35}=\frac{61}{35}\)

\(\frac{-5}{7}\cdot\frac{3}{13}-\frac57\cdot\frac{10}{13}+1\frac57\)

\(=\frac57\cdot\frac{-3}{13}+\frac57\cdot\frac{-10}{13}+1+\frac57\)

\(=\frac57\cdot\left(\frac{-3}{13}+\frac{-10}{13}+1\right)+1\)

\(=\frac57\cdot\left(-1+1\right)+1\)

\(=\frac57\cdot0+1\)

\(=0+1=1\)

- \(\frac57\).\(\frac{3}{13}\) - \(\frac57\).\(\frac{10}{13}\) + 1\(\frac57\)

= -\(\frac57\).(\(\frac{3}{13}+\frac{10}{13}\)) + 1 + \(\frac57\)

= - 5/7. 1 + 5/7 + 1

= -5/7 + 5/7 + 1

= 0 + 1

= 1

4/3 + -11/31 + 3/10 - 20/31 - 2/5

= (4/3 + 3/10 - 2/5) - (11/31 + 20/31)

= (40/30 + 9/30 - 12/30) - 1

= (49/30 - 12/30) - 1

= 37/30 - 30/30

= 7/30

7/30 đúng ko

vì -154/-156 là phần số,phần số -154/-156=154/156
Vì tử số không bằng/lớn hơn mẫu số,nên:
-154/-156<1
Hoặc đổi 1 ra phần số:
-154/-156<156/156(1)
1Lớn hơn do phần tử lớn hơn

23,75 + 18,5 - ( 6,5 - 76,25)

= 23,75 + 18,5 - 6,5 + 76,25

= 42,25 - 6,5 + 76,25

= 35,75 + 76,25

= 112

23,75 + 18,5 - (6,5 - 76,25)

= 23,75 + 18,5 - 6,5 + 76,25

= (23,75 + 76,25) + (18,5 - 6,5)

= 100 + 12

= 112

Giải:

Vận tốc của ca nô khi ngược dòng là:

50 - 5 x 2 = 40(km/h)

30 phút = 0,5 giờ

Gọi thời gian ca nô xuôi dòng là t(giờ); t > 0

Thì thời gian ngược dòng là: t + 0,5 (giờ)

Theo bài ra ta có phương trình:

40.(t + 0,5) = 50.t

40t + 20 = 50t

50t - 40t = 20

10t = 20

t = 20 : 10

t = 2

Thời gian ca nô xuôi dòng là 2 giờ

Quãng sông AB dài là: 50 x 2 = 100(km)

Kết luận:...

Ta có:

\(S = \frac{1}{2 \cdot 3} + \frac{1}{4 \cdot 5} + \frac{1}{6 \cdot 7} + . . . + \frac{1}{2022 \cdot 2023}\)

Với:

\(\frac{1}{n \left(\right. n + 1 \left.\right)} = \frac{1}{n} - \frac{1}{n + 1}\)

Suy ra:

\(S = \left(\right. \frac{1}{2} - \frac{1}{3} \left.\right) + \left(\right. \frac{1}{4} - \frac{1}{5} \left.\right) + \left(\right. \frac{1}{6} - \frac{1}{7} \left.\right) + . . . + \left(\right. \frac{1}{2022} - \frac{1}{2023} \left.\right)\)

Do đó:

\(S = \left(\right. \frac{1}{2} + \frac{1}{4} + \frac{1}{6} + . . . + \frac{1}{2022} \left.\right) - \left(\right. \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + . . . + \frac{1}{2023} \left.\right)\)

Xét từng cặp:

\(\frac{1}{2} > \frac{1}{3} , \&\text{nbsp}; \frac{1}{4} > \frac{1}{5} , \&\text{nbsp}; . . . , \&\text{nbsp}; \frac{1}{2022} > \frac{1}{2023}\)

Suy ra:

\(S > 0\)

Mặt khác:

\(\frac{1}{n \left(\right. n + 1 \left.\right)} < \frac{1}{n^{2}} \Rightarrow S < \frac{1}{2^{2}} + \frac{1}{4^{2}} + . . . < \frac{1}{2}\)

Ta có:

\(\frac{1011}{2023} \approx 0,4997 < \frac{1}{2}\)

Và thực tế tổng \(S\) nhỏ hơn giá trị này.

Kết luận:

\(S < \frac{1011}{2023}\)

\(S=\frac{2}{3 \times5}+\frac{2}{5 \times7}+\frac{2}{7 \times9}+\ldots+\frac{2}{97 \times99}\)

\(S=\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{7}\right)+\left(\frac{1}{7}-\frac{1}{9}\right)+\ldots+\left(\frac{1}{97}-\frac{1}{99}\right)\)

\(S = \frac{1}{3} - \frac{1}{99}\)

\(S = \frac{33}{99} - \frac{1}{99}\)

\(S = \frac{32}{99}\)