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ND
1
HT
13 tháng 7 2015
\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)....\left(\frac{1}{2003}-1\right)\)
=\(\frac{-1}{2}.\frac{-2}{3}.\frac{-3}{4}.....\frac{-2002}{2003}\)
=\(\frac{1}{2003}\)
\(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}....\frac{9999}{10000}\)
=\(\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{99.101}{100.100}\)
=\(\frac{\left(1.2.3.....99\right)\left(3.4.5.....101\right)}{\left(2.3.4.....100\right)\left(2.3.4.....100\right)}\)
=\(\frac{101}{100.2}\)
=\(\frac{101}{200}\)
VC
Vũ Cao Minh⁀ᶦᵈᵒᶫ ( ✎﹏IDΣΛ亗 )
CTV
VIP
16 tháng 7 2019
\(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.....\frac{9999}{10000}\)
\(A=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{99.101}{100.100}\)
\(A=\frac{\left(1.2.3.....99\right).\left(3.4.5.....101\right)}{\left(2.3.4.....100\right).\left(2.3.4.....100\right)}\)
\(A=\frac{1.101}{2.100}=\frac{101}{200}\)
16 tháng 7 2019
\(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}......\frac{9999}{10000}\)
\(A=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{99.101}{100.100}\)
\(A=\frac{1.2.3.4.....99}{2.3.4.5.....100}.\frac{3.4.5.6.....101}{2.3.4.5.....100}\)
\(A=\frac{1}{100}.\frac{101}{2}\)
\(A=\frac{101}{200}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
14 tháng 10
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 = \(\dfrac{101}{200}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
14 tháng 10
2; B = (1 - \(\dfrac{1}{2}\)).(1 - \(\dfrac{1}{8}\))...(1 - \(\dfrac{1}{n+1}\))
Xem lại đề bài.
3 tháng 9 2015
3/4.8/9.15/16......9999/10000
= 3.8.15.....9999/4.9.16......10000
=101/50
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
14 tháng 10
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.\(\dfrac{5}{12}\)
= \(\dfrac{25}{12}\)
25 tháng 8 2021
\(M=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot\cdot\cdot\cdot\frac{9999}{10000}\)
\(=\frac{1.3}{2.2}\cdot\frac{2.4}{3.3}\cdot\frac{3.5}{4.4}\cdot\cdot\cdot\cdot\frac{99.101}{100.100}\)
\(=\frac{1}{2}\cdot\frac{101}{100}=\frac{101}{200}\)
25 tháng 8 2021
Xét vế phải :
\(VP=\frac{99}{50}-\frac{97}{49}+...+\frac{7}{4}-\frac{5}{3}+\frac{3}{2}-1\)
\(=2.\left(\frac{99}{100}-\frac{97}{98}+...+\frac{7}{8}-\frac{5}{6}+\frac{3}{4}-\frac{1}{2}\right)\)
\(=2\left[\left(1-\frac{1}{100}\right)-\left(1-\frac{1}{98}\right)+...+\left(1-\frac{1}{4}\right)-\left(1-\frac{1}{2}\right)\right]\)
\(=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{100}\right)\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{25}+\frac{1}{26}+...+\frac{1}{50}\right)-\left(1+\frac{1}{2}+...+\frac{1}{25}\right)\)
\(=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{49}+\frac{1}{50}=VT\Rightarrow\left(đpcm\right)\)
NC
2
10 tháng 8 2019
\(x=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{9999}{10000}\)
\(x=\frac{1.3}{2.2}+\frac{2.4}{3.3}+\frac{3.5}{4.4}+...+\frac{99.101}{100.100}\)
\(x=\frac{1.2...99}{2.3...100}.\frac{3.4...101}{2.3...100}\)
\(x=\frac{1}{100}.\frac{101}{2}\)
\(x=\frac{101}{200}\)
10 tháng 8 2019
\(X=\frac{1.3}{2.2}+\frac{2.4}{3.3}+\frac{3.5}{4.4}+...+\frac{99.101}{100.100}\)
\(X=\frac{1.2.3....99}{2.3.4....100}.\frac{3.4.5....101}{2.3.4....100}\)
\(X=\frac{1}{100}.\frac{101}{2}\)
\(X=\frac{101}{200}\)
Study well
LM
0
2H
1
LM
1
AH
Akai Haruma
Giáo viên
16 tháng 7 2023
Lời giải:
$A=(1-\frac{1}{4})+(1-\frac{1}{9})+(1-\frac{1}{16})+....+(1-\frac{1}{10000})$
$=(1+1+...+1)-(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+....+\frac{1}{10000})$
$=99-(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+....+\frac{1}{10000})< 99$
\(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...\frac{9999}{10000}\)
\(A=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{99.101}{100.100}\)
\(A=\frac{1.2.3...99}{2.3.4...100}.\frac{3.4.5...101}{2.3.4...100}=\frac{1}{100}.\frac{101}{2}=\frac{101}{200}\)