\(M=\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.....\frac{10^2}{10.11}\)

\(M=\frac{1.1}{1.2}.\frac{2.2}{2.3}.\frac{3.3}{3.4}......\frac{10.10}{10.11}\)

\(M=\frac{1.2.3.....10}{1.2.3....10}.\frac{1.2.3.....10}{2.3.4.....11}\)

\(M=1.\frac{1}{11}\)

\(M=\frac{1}{11}\)

Cảm ơn bạn nha

\(\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.....\frac{10^2}{10.11}\)

\(=\frac{1.1}{1.2}.\frac{2.2}{2.3}.\frac{3.3}{3.4}......\frac{10.10}{10.11}\)

\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{10}{11}\)

\(=\frac{1.2.3.....10}{2.3.4.....11}=\frac{1}{11}\)

a) \(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}< 1\)

\(\Rightarrow A< 1\)

b) \(B=\frac{1}{3}+\left(\frac{1}{3}\right)^2+...+\left(\frac{1}{3}\right)^{100}\)

\(\Rightarrow3B=1+\frac{1}{3}+...+\left(\frac{1}{3}\right)^{99}\)

\(\Rightarrow3B-B=1-\left(\frac{1}{3}\right)^{100}\)

\(\Rightarrow2B=1-\left(\frac{1}{3}\right)^{100}< 1\)

\(\Rightarrow2B< 1\)

\(\Rightarrow B< \frac{1}{2}\)

c)\(C=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(=\left(1+\frac{1}{3}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}-1-\frac{1}{2}-...-\frac{1}{25}\)

\(=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\)

Ta có: \(\frac{1}{2}+\frac{1}{3}< 2\cdot\frac{1}{2}=1\)

\(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}< 4\cdot\frac{1}{4}=1\)

\(\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+...+\frac{1}{15}< 8\cdot\frac{1}{8}=1\)

\(\frac{1}{16}+\frac{1}{17}+\frac{1}{18}+...+\frac{1}{31}< 16\cdot\frac{1}{16}=1\)

\(\frac{1}{32}+\frac{1}{33}+\frac{1}{34}+...+\frac{1}{63}< 32\cdot\frac{1}{32}=1\)

Cộng từng vế của các BĐT trên ta có:

\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{63}< 5\)

\(\Leftrightarrow64+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{63}< 69\)

\(\Leftrightarrow1+\frac{1}{1}+1+\frac{1}{2}+1+\frac{1}{3}+...+1+\frac{1}{63}< 69\)

\(\Leftrightarrow\frac{2}{1}+\frac{3}{2}+\frac{4}{3}+...+\frac{64}{63}< 69\)

\(\Leftrightarrow\frac{2^2}{1\cdot2}+\frac{3^2}{2\cdot3}+\frac{4^2}{3\cdot4}+...+\frac{64^2}{63\cdot64}< 69\)đpcm

Cho Linh xin 2 k nào :D

1-1/2+1/2-1/3+1/3-1/4+...+1/9-1/10

= 1-1/10

=9/10

bn trân ơi, sao ra 1- (1/10) hay zậy?

1/

+) \(\frac{3}{6}=\frac{2}{4};\frac{3}{2}=\frac{6}{4};\frac{4}{6}=\frac{2}{3};\frac{4}{2}=\frac{6}{3}\)

2/

\(A=\frac{3n-5}{n+4}=\frac{3n+12-17}{n+4}=\frac{3\left(n+4\right)}{n+4}-\frac{17}{n+4}=3-\frac{17}{n+4}\)

Để A nguyên <=> n + 4 thuộc Ư(17) = {1;-1;17;-17}

Vậy...

3/

\(S=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2016.2017}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2016}-\frac{1}{2017}\)

\(=1-\frac{1}{2017}\)

\(=\frac{2016}{2017}\)

\(A=\frac{3n+12-7}{n+4}=\frac{3\left(n+4\right)}{n+4}-\frac{7}{n+4}=3-\frac{7}{n+4}\)

=> n-4 \(\in\) Ư (7)

n-4=1

n=4+1=5

n-4=-1

n=-1+4=3

n-4=7

n=4+7=11

n-4=-7

n=-7+4=-3

\(S=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2015.2016}+\frac{1}{2016.2017}\)

  \(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2016}-\frac{1}{2017}\)

  \(=1-\frac{1}{2017}\)

  \(=\frac{2016}{2017}\)

\(\frac{1.1}{1.2}.\frac{2.2}{2.3}\frac{3.3}{3.4}...\frac{100.100}{100.101}\)

\(=\frac{\left(1.2.3...100\right).\left(1.2.3...100\right)}{\left(1.2.3...100\right).\left(2.3...101\right)}\)

\(=\frac{1}{1.101}\)

\(=\frac{1}{101}\)

k cho mk nha

dấu bằng của mk bt liệt nên bạn thông cảm 

A bằng (1.2.3.4).(1.2.3.4)/(1.2.3.4).(2.3.4.5) bằng 5 

rút gọn cho nhau bạn nhé

thank bạn nhóa