=A*(1/100-1/10^2)

=0

\(=\left(\dfrac{1}{100}-\dfrac{1}{1^2}\right)\left(\dfrac{1}{100}-\dfrac{1}{4}\right)\cdot...\cdot\left(\dfrac{1}{100}-\dfrac{1}{10^2}\right)\cdot...\cdot\left(\dfrac{1}{100}-\dfrac{1}{400}\right)\)

\(=\left(\dfrac{1}{100}-\dfrac{1}{100}\right)\cdot\left(\dfrac{1}{100}-1\right)\cdot...\cdot\left(\dfrac{1}{100}-\dfrac{1}{400}\right)\)

\(=0\cdot\left(\dfrac{1}{100}-1\right)\cdot...\cdot\left(\dfrac{1}{100}-\dfrac{1}{400}\right)=0\)

\(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{100}\)

\(=\left(1+\frac{1}{100}\right)+\left(\frac{1}{2}+\frac{1}{99}\right)+....+\left(\frac{1}{50}+\frac{1}{51}\right)\)

\(=\frac{101}{1.100}+\frac{101}{2.99}+....+\frac{101}{50.51}\)

\(=101.\left(\frac{1}{1.100}+\frac{1}{2.99}+...+\frac{1}{50.51}\right)\)

Vế mẫu :

 \(\frac{1}{1.100}+\frac{1}{2.99}+......+\frac{1}{1.100}\)

\(=2\left(\frac{1}{1.100}+\frac{1}{2.99}+....+\frac{1}{50.51}\right)\)

Vậy kết quả là :

 \(\frac{101}{2}\)

Tử số = 1 + 1/2 + 1/3 + 1/4 + ... + 1/100

= (1 + 1/100) + (1/2 + 1/99) + ... + (1/50 + 1/51)

= 101/1.100 + 101/2.99 + ... + 101/50.51

= 101.(1/1.100 + 1/2.99 + ... + 1/50.51)

Mẫu số = 1/1.100 + 1/2.99 + 1/3.98 + ... + 1/99.2 + 1/100.1

= 2.(1/1.100 + 1/2.99 + ... + 1/50.51)

=> phân số đề bài cho = 101/2

có thể biến đổi mẫu số cho mình hông 

c: \(100C=\dfrac{100^{100}+100}{100^{100}+1}=1+\dfrac{99}{100^{100}+1}\)

\(100D=\dfrac{100^{101}+100}{100^{101}+1}=1+\dfrac{99}{100^{101}+1}\)

100^100+1<100^101+1

=>\(\dfrac{99}{100^{100}+1}>\dfrac{99}{100^{101}+1}\)

=>100C>100D

=>C>D

b: \(2020E=\dfrac{2020^{2022}+2020}{2020^{2022}+1}=1+\dfrac{2019}{2020^{2022}+1}\)

\(2020F=\dfrac{2020^{2021}+2020}{2020^{2021}+1}=1+\dfrac{2019}{2020^{2021}+1}\)

2020^2022+1>2020^2021+1(Do 2022>2021)

=>\(\dfrac{2019}{2020^{2022}+1}< \dfrac{2019}{2020^{2021}+1}\)

=>2020E<2020F

=>E<F

A = (\(\dfrac{1}{100}\) - 1²).(\(\dfrac{1}{100}\) - \(\dfrac{1}{2^2}\)).(\(\dfrac{1}{100}\) - \(\dfrac{1}{3^2}\))...(\(\dfrac{1}{100}\) - \(\dfrac{1}{20^2}\))

A = (\(\dfrac{1}{10^2}\) - 1²).(\(\dfrac{1}{10^2}\) - \(\dfrac{1}{2^2}\)).(\(\dfrac{1}{10^2}\) - \(\dfrac{1}{3^2}\))..(\(\dfrac{1}{10^2}\) - \(\dfrac{1}{10^2}\))....(\(\dfrac{1}{10^2}\) - \(\dfrac{1}{20^2}\))

A = (\(\dfrac{1}{10^2}\) - 1²).(\(\dfrac{1}{10^2}\) - \(\dfrac{1}{2^2}\)).(\(\dfrac{1}{10^2}\) - \(\dfrac{1}{3^2}\))...0.(\(\dfrac{1}{10^2}\) - \(\dfrac{1}{20^2}\))

A = 0

bạn viết sai đề rồi

dung de roi ma

a. Có: \(\frac{100^{101}+1}{100^{100}+1}>1\Rightarrow\frac{100^{101}+1}{100^{100}+1}>\frac{100^{101}+\left(1+99\right)}{100^{100}+\left(1+99\right)}\)

\(\Rightarrow B>\frac{100^{101}+100}{100^{100}+100}\\ \Rightarrow B>\frac{100\left(100^{100}+1\right)}{100\left(100^{99}+1\right)}\\ \Rightarrow B>\frac{100^{100}+1}{100^{99}+1}=A\\ \Leftrightarrow A< B\)

Vậy A < B

b. Có: \(\frac{13^{16}+1}{13^{17}+1}< 0\Rightarrow\frac{13^{16}+1}{13^{17}+1}< \frac{13^{16}+\left(1+12\right)}{13^{17}+\left(1+12\right)}\)

\(\Rightarrow B< \frac{13^{16}+13}{13^{17}+13}\\ \Rightarrow B< \frac{13\left(13^{15}+1\right)}{13\left(13^{16}+1\right)}\\ \Rightarrow B< \frac{13^{15}+1}{13^{16}+1}=A\\ \Leftrightarrow A>B\)

Vậy A > B

c. Có: \(\frac{1999^{2000}+1}{1999^{1999}+1}>1\Rightarrow\frac{1999^{2000}+1}{1999^{1999}+1}>\frac{1999^{2000}+\left(1+1998\right)}{1999^{1999}+\left(1+1998\right)}\)

\(\Rightarrow B>\frac{1999^{2000}+1999}{1999^{1999}+1999}\\ \Rightarrow B>\frac{1999\left(1999^{1999}+1\right)}{1999\left(1999^{1998}+1\right)}\\ \Rightarrow B>\frac{1999^{1999}+1}{1999^{1998}+1}=A\\ \Leftrightarrow A< B\)

Vậy A < B

tu 1 den 100 co 100 so

nen tong cac so do la : ( 100 + 1 ) x 100 : 2 = 5050

nhin tong tren , ta thay moi so duoc lap lai 4 lan nen tong do la : 5050 x 4 = 20200 

                        dap so : 20200

Tính có bao nhiêu số hạng: (100-1):1+1 x 5= 500(số)

Tính tổng của dãy số trên: (100+10) x 500 :2 x 5 =137500

tổng = 5050 vì mỗi số xh 4 lần nên tg = 5050*4=20200