Mày hay nhờ mai tao méc thầy

tự làm đi

\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)....\left(\frac{1}{2003}-1\right)\)

=\(\frac{-1}{2}.\frac{-2}{3}.\frac{-3}{4}.....\frac{-2002}{2003}\)

=\(\frac{1}{2003}\)

\(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}....\frac{9999}{10000}\)

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

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

=\(\frac{101}{100.2}\)

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

\(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.....\frac{9999}{10000}\)

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

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

\(A=\frac{1.101}{2.100}=\frac{101}{200}\)

\(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}......\frac{9999}{10000}\)

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

\(A=\frac{1.2.3.4.....99}{2.3.4.5.....100}.\frac{3.4.5.6.....101}{2.3.4.5.....100}\)

\(A=\frac{1}{100}.\frac{101}{2}\)

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

\(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...\frac{9999}{10000}\)

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

\(A=\frac{1.2...99}{2.3...100}.\frac{3.5...101}{2.3...100}\)

\(A=\frac{1}{100}.\frac{101}{2}\)

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

\(B=\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)\left(1-\frac{1}{15}\right)...\left(1-\frac{1}{780}\right)\)

\(B=\frac{2}{3}.\frac{5}{6}.\frac{9}{10}.\frac{14}{15}...\frac{779}{780}\)

\(B=\frac{4}{6}.\frac{10}{12}.\frac{18}{20}....\frac{1558}{1560}\)

\(B=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}...\frac{38.41}{39.40}\)

\(B=\frac{1.2.3...38}{2.3...39}.\frac{4.5...41}{3.4...40}\)

\(B=\frac{1}{39}.\frac{41}{3}\)

\(B=\frac{41}{117}\)

\(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}....\frac{9999}{10000}\)

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

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

\(=\frac{1}{100}.\frac{101}{2}=\frac{101}{200}\)

\(B=\left(1-\frac{1}{4}\right).\left(1-\frac{1}{9}\right).....\left(1-\frac{1}{10000}\right)\)

\(=\frac{3}{4}.\frac{8}{9}....\frac{9999}{10000}=\frac{101}{200}\)

A = \(\dfrac{3}{4}\).\(\dfrac{8}{9}\).\(\dfrac{15}{16}\)...\(\dfrac{9999}{10000}\)

A = \(\dfrac{1.3.2.4..3.5......99.101}{2.2.3.3.4.4....100.100}\)

A = \(\dfrac{1.2.3..4.5.....99}{2.3.4.5.....99.100}\).\(\dfrac{3.4.5....100.101}{2.3.4.5...100}\)

A = \(\dfrac{1}{100}\).\(\dfrac{101}{2}\)

A = \(\dfrac{101}{200}\)

2; B = (1 - \(\dfrac{1}{2}\)).(1 - \(\dfrac{1}{8}\))...(1 - \(\dfrac{1}{n+1}\))

   Xem lại đề bài.

A = \(\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot\cdot\cdot\cdot\frac{9999}{10000}=\frac{1\cdot3}{2.2}\cdot\frac{2\cdot4}{3\cdot3}\cdot\frac{3.5}{4.4}\cdot\cdot\cdot\cdot\frac{99\cdot101}{100\cdot100}=\frac{1}{2}\cdot\frac{101}{100}=\frac{101}{200}\)

B = ( 1- 1/4 )( 1-1/9) ...( 1-1/10000 ) = 3/4 . 8/9 .....9999/100000 ( tương tự A )

a=5051/100 co ma

xin loi ban lam dung roi minh nham

3/4.8/9.15/16......9999/10000
= 3.8.15.....9999/4.9.16......10000
=101/50

a; \(\dfrac{5}{6}\) + \(\dfrac{5}{12}\) + \(\dfrac{5}{20}\) + ... + \(\dfrac{5}{132}\)

 = 5.(\(\dfrac{1}{6}\) + \(\dfrac{1}{12}\) + \(\dfrac{1}{20}\) + ..+ \(\dfrac{1}{132}\))

= 5.(\(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + ... + \(\dfrac{1}{11.12}\))

= 5.(\(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + ...+ \(\dfrac{1}{11}\) - \(\dfrac{1}{12}\))

= 5.(\(\dfrac{1}{2}\) - \(\dfrac{1}{12}\))

= 5.(\(\dfrac{6}{12}\) - \(\dfrac{1}{12}\))

= 5.\(\dfrac{5}{12}\)

= \(\dfrac{25}{12}\)

b; \(\dfrac{1}{2.4}\) + \(\dfrac{1}{4.6}\) + ... + \(\dfrac{1}{x.\left(x+2\right)}\)

= \(\dfrac{1}{2}\) .(\(\dfrac{2}{2.4}\) + \(\dfrac{2}{4.6}\) + ... + \(\dfrac{2}{x\left(x+2\right)}\))

= \(\dfrac{1}{2}\).(\(\dfrac{1}{2}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{6}\) +...+ \(\dfrac{1}{x}\) - \(\dfrac{1}{x+2}\))

= \(\dfrac{1}{2}\).(\(\dfrac{1}{2}\) - \(\dfrac{1}{x+2}\))

= \(\dfrac{1}{2}\).\(\dfrac{x+2-2}{x+2}\)

= \(\dfrac{x+\left(2-2\right)}{2.\left(x+2\right)}\)

= \(\dfrac{x+0}{2\left(x+2\right)}\)

= \(\dfrac{x}{2.\left(x+2\right)}\)

3/4.8/9.15/16.....624/625

=(1.3)/(2.2).(2.4)/(3.3).(3.5)/(4.4)...(24.26)/(25.25)

=(1.2.3....24).(3.4.5....26)/(2.3.4...25).(2.3.4...25)

=26/25.2

=26/50

=13/25

Đặt\(A=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{300}\)

Vì\(\frac{1}{101}>\frac{1}{102}>\frac{1}{103}>...>\frac{1}{300}\)

\(\Rightarrow\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\right)+\left(\frac{1}{201}+\frac{1}{202}+...+\frac{1}{300}\right)\)\(>\left(\frac{1}{200}+\frac{1}{200}+...+\frac{1}{200}\right)+\left(\frac{1}{300}+\frac{1}{300}+...+\frac{1}{300}\right)\)(mỗi cái trong ngoặc là một trăm phân số)

\(\Rightarrow\frac{1}{101}+\frac{1}{102}+...+\frac{1}{300}>\left(\frac{1}{200}\right).100+\left(\frac{1}{300}\right).100\)

\(\Rightarrow A>\frac{1}{2}+\frac{1}{3}\)

\(\Rightarrow A>\frac{5}{6}\)

Mà 5/6>2/3=>A>2/3

\(\Rightarrow\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{300}\)

Đặt A = \(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{300}\)

Vì \(\frac{1}{101}>\frac{1}{102}>\frac{1}{103}>...>\frac{1}{300}\)

\(\Rightarrow\left(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+....\frac{1}{200}\right)+\left(\frac{1}{201}+\frac{1}{202}+\frac{1}{103}+.....\frac{1}{300}\right)>\left(\frac{1}{200}+\frac{1}{200}+\frac{1}{200}\right)\)

Tự làm tiếp nhé !!!