\(=1\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)

\(=1\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{9}-\frac{1}{10}\right)\)

\(=1-\frac{1}{10}\)

\(=\frac{9}{10}\)

1/2-1/10=2/5

Chúc bạn may mắn@

\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)

\(=\frac{1}{2}-\frac{1}{10}\)

=\(\frac{2}{5}\)

k nha!

\(=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(\Rightarrow\frac{1}{2}-0+0+...+0-\frac{1}{100}\)

\(\Rightarrow\frac{50}{100}-\frac{1}{100}=\frac{49}{100}\)

\(A=\frac{5}{1.2}+\frac{5}{2.3}+...+\frac{5}{7.8}\)

\(\Rightarrow5A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{7.8}\)

\(\Rightarrow5A=1.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{8}\right)\)

\(\Rightarrow5A=1-\frac{1}{8}\)

\(\Rightarrow A=\left(1-\frac{1}{8}\right).\frac{1}{5}=\frac{7}{40}\)

\(A=\frac{5}{1.2}+\frac{5}{2.3}+...+\frac{5}{7.8}\)

\(A=5\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{5}{7.8}\right)\)

\(A=5\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{8}\right)\)

\(A=5\left(1-\frac{1}{8}\right)\)

\(A=5.\frac{7}{8}\)

\(A=\frac{38}{8}\)

nhầm rồi :)

\(A=5.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-....-\frac{1}{8}\right)\)

\(\Rightarrow A=\frac{5.7}{8}\)

\(\Rightarrow A=\frac{35}{8}\)

\(1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\)

\(=1+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\right)\)

\(=1+\left(1-\frac{1}{2018}\right)\)

\(=1+\left(\frac{2018}{2018}-\frac{1}{2018}\right)\)

\(=1+\left(\frac{2017}{2018}\right)\)

\(=\frac{2018}{2018}+\frac{2017}{2018}=\frac{4035}{2018}\)

\(1+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}...+\frac{1}{2017\cdot2018}\)

\(=1+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}...+\frac{1}{2017}-\frac{1}{2018}\right)\)

\(=1+\left(1-\frac{1}{2018}\right)\)

\(=1+\frac{2017}{2018}\)

\(=1+\frac{2017}{2018}\)

\(=\frac{4035}{2018}\)

\(1+\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{2017x2018}\)

\(=1+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\right)\)

\(=1+\left(1-\frac{1}{2018}\right)\)

\(=1+\frac{2017}{2018}\)

\(=\frac{4035}{2018}\)

\(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}\)

\(=\frac{3-2}{2\times3}+\frac{4-3}{3\times4}+\frac{5-4}{4\times5}+\frac{6-5}{5\times6}\)

\(=\frac{3}{2\times3}-\frac{2}{2\times3}+\frac{4}{3\times4}-\frac{3}{3\times4}+\frac{5}{4\times5}-\frac{4}{4\times5}+\frac{6}{5\times6}-\frac{5}{5\times6}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)

\(=\frac{1}{2}-\frac{1}{6}\)

\(=\frac{1}{3}\)

\(G=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)

\(G=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^5}\)

\(3G=3+1+\frac{1}{3}+...+\frac{1}{3^4}\)

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

\(2G=3-\frac{1}{3^5}\)

\(2G=3-\frac{1}{243}\)

\(2G=\frac{729}{243}-\frac{1}{243}\)

\(G=\frac{728}{243}:2\)

\(G=\frac{364}{243}\)

\(\frac{3}{1.2}+\frac{3}{2.3}+...+\frac{3}{x.\left(x+1\right)}=\frac{6042}{2015}\)

\(3.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{6042}{2015}\)

\(1-\frac{1}{x+1}=\frac{6042}{2015}:3\)

\(1-\frac{1}{x-1}=\frac{2014}{2015}\)

\(\frac{1}{x-1}=1-\frac{2014}{2015}\)

\(\frac{1}{x-1}=\frac{1}{2015}\)

\(\Rightarrow x-1=2015\)

\(\Rightarrow x=2016\)

\(\frac{3}{2\times3}\)+\(\frac{3}{3x4}\)+\(\frac{3}{4x5}\)+ ... +\(\frac{3}{96x97}\)

= \(\frac{3}{2}\)-\(\frac{3}{3}\)+ \(\frac{3}{3}\)- \(\frac{3}{4}\)+\(\frac{3}{4}\)-\(\frac{3}{5}\)+ ... + \(\frac{3}{96}\)- \(\frac{3}{97}\)

ở giữa cứ trù \(\frac{3}{3}\) rồi lại cộng \(\frac{3}{3}\)thì hết nên cụm ở giữa là hết

chỉ còn \(\frac{3}{2}\)-\(\frac{3}{97}\)= \(\frac{285}{194}\)

vậy đáp án câu này là \(\frac{285}{194}\)

/HT\

a) Đặt \(A=\frac{1^2}{1.2}+\frac{2^2}{2.3}+.........+\frac{100^2}{100.101}\)

\(\Rightarrow A=\left(1^2+2^2+..........+100^2\right)\)\(.\left(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{100.101}\right)\)

\(\Rightarrow A=\left(1^2+2^2+......+100^2\right).\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{100}-\frac{1}{101}\right)\)

\(\Rightarrow A=\left(1^2+2^2+......+100^2\right).\left(1-\frac{1}{101}\right)\)

\(\Rightarrow A=\left(1^2+2^2+.....+100^2\right).\left(\frac{100}{101}\right)\)^(a)

Đặt \(M=\left(1^2+2^2+........+100^2\right)\)

\(\Rightarrow M=1.1+2.2+.....+100.100\)

\(\Rightarrow M=1.\left(2-1\right)+2.\left(3-1\right)+....+100.\left(101-1\right)\)

\(\Rightarrow M=\left(1.2-1\right)+\left(2.3-2\right)+.....+\left(100.101-100\right)\)

\(\Rightarrow M=\left(1.2+2.3+.....+100.101\right)-\left(1+2+......+100\right)\)

\(\Rightarrow M=\left(1.2+2.3+......+100.101\right)-5050\)⁽¹⁾

Đặt \(N=1.2+2.3+....+100.101\)

\(\Rightarrow3.N=1.2.3+2.3.3+......+100.101.3\)

\(\Rightarrow3N=1.2.\left(3-0\right)+2.3.\left(4-1\right)+......+100.101.\left(102-99\right)\)

\(\Rightarrow3N=\left(1.2.3-0\right)+\left(1.2.3-2.3.4\right)+.......+\left(100.101.102-100.101.99\right)\)

\(\Rightarrow3N=100.101.102-0\)

\(\Rightarrow N=343400\)

Thay N = 343400 vào 1) ta được:

M = 343400 - 5050 

=> M = 338350

Thay M = 338350 Vào (a) ta được:

A = 338350 . \(\frac{100}{101}\)

=> \(A=\frac{33835000}{101}\)

Vậy \(\frac{1^2}{1.2}+\frac{2^2}{2.3}+.........+\frac{100^2}{100.101}=\frac{33835000}{101}=335000\)

b) Đặt \(B=\frac{2^2}{1.3}+\frac{3^2}{2.4}+..........+\frac{59^2}{58.60}\)

\(\Rightarrow B=\left(2^2+3^2+........+59^2\right).\left(\frac{1}{1.3}+\frac{1}{2.4}+.....+\frac{1}{58.60}\right)\)

Đặt \(G=2^2+3^2+.........+59^2\)VÀ \(H=\frac{1}{1.3}+\frac{1}{2.4}+.........+\frac{1}{58.60}\)

\(\Rightarrow G=2.2+3.3+.......+59.59\) VÀ \(2.H=\frac{2}{1.3}+\frac{2}{2.4}+......+\frac{2}{58.60}\)

Rồi bạn làm như ở phần a) ý