a)\(A=\frac{31}{23}-\left(\frac{7}{32}+\frac{8}{2}\right)vaB=\left(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}\right)-\left(\frac{79}{67}-\frac{28}{41}\right)\)

+)Ta có:\(A=\frac{31}{23}-\left(\frac{7}{32}+\frac{8}{2}\right)\)

\(\Leftrightarrow A=\frac{31}{23}-\left(\frac{7}{32}+\frac{128}{32}\right)\)

\(\Leftrightarrow A=\frac{31}{23}-\frac{135}{32}\)

\(\Leftrightarrow A=\frac{992}{736}-\frac{3105}{736}\)

\(\Leftrightarrow A=\frac{-2113}{736}\left(1\right)\)

+)Ta lại có:\(B=\left(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}\right)-\left(\frac{79}{67}-\frac{28}{41}\right)\)

\(\Leftrightarrow B=\frac{1}{3}+\frac{12}{67}+\frac{13}{41}-\frac{79}{67}+\frac{28}{41}\)

\(\Leftrightarrow B=\frac{1}{3}+\left(\frac{12}{67}-\frac{79}{67}\right)+\left(\frac{13}{41}+\frac{28}{41}\right)\)

\(\Leftrightarrow B=\frac{1}{3}+\frac{-67}{67}+\frac{41}{41}\)

\(\Leftrightarrow B=\frac{1}{3}+\left(-1\right)+1\)

\(\Leftrightarrow B=\frac{1}{3}\left(2\right)\)

+)Từ (1) và (2) 

\(\Leftrightarrow A< 0< B\Leftrightarrow A< B\)

Vậy A<B

b)\(\frac{200420042004}{200520052005}va\frac{2004}{2005}\)

+)Ta có \(\frac{200420042004}{200520052005}=\frac{2004.100010001}{2005.100010001}=\frac{2004}{2005}\)

\(\Leftrightarrow\frac{200420042004}{200520052005}=\frac{2004}{2005}\)

c)\(C=\frac{2020^{2006}+1}{2020^{2007}+1}vaD=\frac{2020^{2005}+1}{2020^{2006}+1}\)

\(C=\frac{2020^{2006}+1}{2020^{2007}+1}< 1\)

\(\Leftrightarrow C< \frac{2020^{2006}+1+2019}{2020^{2007}+1+2019}=\frac{2020^{2006}+2020}{2020^{2007}+2020}=\frac{2020.\left(2020^{2005}+1\right)}{2020.\left(2020^{2006}+1\right)}=\frac{2020^{2005}+1}{2020^{2006}+1}\)

\(\Leftrightarrow C< D\)

Chúc bạn học tốt

\(4\cdot5^{100}\cdot\left(\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{100}}\right)+1\)

\(=4\cdot\left(\frac{5^{100}}{5}+\frac{5^{100}}{5^2}+\frac{5^{100}}{5^3}+...+\frac{5^{100}}{5^{100}}\right)+1\)

\(=4\cdot\left(5^{99}+5^{98}+5^{97}+...+1\right)+1\)

\(\text{Đặt }S=5^{99}+5^{98}+5^{97}+...+1\)

\(5S=5^{100}+5^{99}+5^{98}+...+5\)

\(5S-S=5^{100}-4\)

\(4S=5^{100}-4\)

\(S=\frac{5^{100}-4}{4}\)

\(\text{Quay lại bài toán ta có : }\)

 \(4\cdot\left(\frac{5^{100}}{5}+\frac{5^{100}}{5^2}+\frac{5^{100}}{5^3}+...+\frac{5^{100}}{5^{100}}+1=\right)\)        \(4\cdot\left(\frac{5^{100}-4}{4}\right)+1\)

\(=5^{100}-4+1\)

\(=5^{100}-3\)

                                  \(\text{Mình nghĩ chắc cách làm này đúng rồi đó ! Bạn tham khảo nha ! Bài mình tự nghĩ đó ! Nếu có sai sót gì bạn tự chỉnh nha !}\)

bn giải thích cho mk đoạn \(5S-S=5^{100}-4\)đc ko sao lại trừ 4

Ta có : \(A=\frac{1}{1.101}+\frac{1}{2.202}+\frac{1}{3.103}+...+\frac{1}{10.110}\)

=\(\frac{1}{100}.\left(\frac{100}{1.101}+\frac{100}{2.102}+\frac{100}{3.103}+...+\frac{100}{10.110}\right)\)

= \(\frac{1}{100}\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+\frac{1}{3}-\frac{1}{103}+...+\frac{1}{10}-\frac{1}{110}\right)\)

= \(\frac{1}{100}\cdot\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}-\frac{1}{101}-\frac{1}{102}-...-\frac{1}{110}\right)\)

Lại có : B = \(\frac{1}{10}.\left(\frac{10}{1.11}+\frac{10}{2.12}+\frac{10}{3.13}+...+\frac{10}{100.110}\right)\)

= \(\frac{1}{10}\left(1-\frac{1}{11}+\frac{1}{2}-\frac{1}{12}+\frac{1}{3}-\frac{1}{13}+...+\frac{1}{100}-\frac{1}{110}\right)\)

= \(\frac{1}{10}\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}-\frac{1}{101}-\frac{1}{102}-...-\frac{1}{110}\right)\)

Khi đó \(A:B=\frac{A}{B}=\frac{\frac{1}{100}\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}-\frac{1}{101}-\frac{1}{102}-...-\frac{1}{110}\right)}{\frac{1}{10}\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}-\frac{1}{101}-\frac{1}{102}-...-\frac{1}{110}\right)}=\frac{1}{10}\)

\(=-\left(1-\frac{1}{2^2}\right).\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)....\left(1-\frac{1}{100^2}\right)\)

\(=-\left(\frac{1.3}{2.2}\right)\left(\frac{2.4}{3.3}\right)\left(\frac{3.5}{4.4}\right)....\left(\frac{99.101}{100.100}\right)\)

\(=-\left(\frac{1.2.3...99}{2.3.4...100}\right)\left(\frac{3.4.5...101}{2.3.4...100}\right)\)

\(=-\left(\frac{1}{100}\right).\left(\frac{101}{2}\right)\)

\(=-\frac{101}{200}\)

K = -3/4.-8/9......-9999/10000

= -(3/4.8/9....9999/10000)

= -(1.3.2.4.....99.101/2^2.3^2.....100^2)

= -(1.2.3.....101).(3.4.5....99)/(2.3.4.....100).(2.3.4....100)

= -(101/2.100)

= -101/200

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

\(2S=\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}+\frac{1}{101}\)

\(\Leftrightarrow2S-S=\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{101}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)

\(\Leftrightarrow S=\frac{1}{101}-1=-\frac{100}{101}\)

Đặt D = 1 + 1/2 + 1/3 + . . . . . + 1/100

Ta có : D = (2 - 1) + 1/2  - 1/3 + 1/3 - 1/4 + . . . . . + 1/99 - 1/100

            D = 1 - 1/100

            D = 100/100 - 1/100

            D = 99/100