B)2/3:1/2-1/4.8
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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 :
\(VP=\frac{99}{50}-\frac{97}{49}+...+\frac{7}{4}-\frac{5}{3}+\frac{3}{2}-1\)
\(=2.\left(\frac{99}{100}-\frac{97}{98}+...+\frac{7}{8}-\frac{5}{6}+\frac{3}{4}-\frac{1}{2}\right)\)
\(=2\left[\left(1-\frac{1}{100}\right)-\left(1-\frac{1}{98}\right)+...+\left(1-\frac{1}{4}\right)-\left(1-\frac{1}{2}\right)\right]\)
\(=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{100}\right)\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{25}+\frac{1}{26}+...+\frac{1}{50}\right)-\left(1+\frac{1}{2}+...+\frac{1}{25}\right)\)
\(=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{49}+\frac{1}{50}=VT\Rightarrow\left(đpcm\right)\)
TT
3
XO
11 tháng 6 2020
\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{1000}\right)=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{999}{1000}=\frac{1.2.3...999}{2.3.4...1000}=\frac{1}{1000}\)
\(B=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}....\frac{2499}{2500}=\frac{3.8.15...2499}{4.9.16....2500}=\frac{1.3.2.4.3.5....49.51}{2.2.3.3.4.4...50.50}=\frac{\left(1.2.3...49\right).\left(3.4.5...51\right)}{\left(2.3.4...50\right).\left(2.3.4...50\right)}\)
\(\frac{1.51}{50.2}=\frac{51}{100}\)
11 tháng 6 2020
a. \(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{999}\right)\)
\(A=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot....\cdot\frac{998}{999}\)
\(A=\frac{1\cdot2\cdot3\cdot....\cdot998}{2\cdot3\cdot4\cdot....\cdot999}=\frac{1}{999}\)
Vậy \(A=\frac{1}{999}\)
16 tháng 9 2018
a) Có \(\dfrac{2^8.9^2}{6^4.8^2}\)=\(\dfrac{2^8.3^4}{3^4.2^42^6}\)=\(\dfrac{2^8.3^4}{3^4.2^{10}}\)=\(\dfrac{2^8}{2^{10}}\)=\(\dfrac{1}{4}\)
b) Có \(\left(2^{-1}+3^{-1}\right):\left(2^{-1}-3^{-1}\right)+\left(2^{-1}.2^0\right).2^3\)
= \(\dfrac{2^{-1}+3^{-1}}{2^{-1}-3^{-1}}+2^{-1}.2^3\)
=\(\dfrac{\dfrac{1}{2}+\dfrac{1}{3}}{\dfrac{1}{2}-\dfrac{1}{3}}+\dfrac{1}{2}.2^3\)
= \(\dfrac{\dfrac{5}{6}}{\dfrac{1}{6}}+2^2\)=5+8=13
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.
AH
Akai Haruma
Giáo viên
14 tháng 9
Lời giải:
\(A=\frac{-1}{2}.\frac{-2}{3}.\frac{-3}{4}....\frac{-998}{999}.\frac{-999}{1000}\\
=\frac{(-1)(-2)(-3)...(-998)(-999)}{2.3.4....1000}\\
=-\frac{1.2.3.4....998.999}{2.3.4...1000}\\
=-\frac{1}{1000}\)
AH
Akai Haruma
Giáo viên
14 tháng 9
Trong $B$ có một thừa số là $1-\frac{7}{7}=0$ nên $B=0$ (do số nào nhân với $0$ cũng sẽ bằng $0$.
----------------------
$C=\frac{1.3}{2^2}.\frac{2.4}{3^2}.\frac{3.5}{4^2}...\frac{49.51}{50^2}$
$=\frac{1.3.2.4.3.5.....49.51}{2^2.3^2.4^2....50^2}$
$=\frac{(1.2.3...49)(3.4.5...51)}{(2.3.4...50)(2.3.4...50)}$
$=\frac{1.2.3...49}{2.3.4...50}.\frac{3.4.5...51}{2.3.4....50}$
$=\frac{1}{50}.\frac{51}{2}=\frac{51}{100}$
ND
1
HT
13 tháng 7 2015
\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)....\left(\frac{1}{2003}-1\right)\)
=\(\frac{-1}{2}.\frac{-2}{3}.\frac{-3}{4}.....\frac{-2002}{2003}\)
=\(\frac{1}{2003}\)
\(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}....\frac{9999}{10000}\)
=\(\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{99.101}{100.100}\)
=\(\frac{\left(1.2.3.....99\right)\left(3.4.5.....101\right)}{\left(2.3.4.....100\right)\left(2.3.4.....100\right)}\)
=\(\frac{101}{100.2}\)
=\(\frac{101}{200}\)
NY
2
H
31 tháng 3 2019
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
\(\Rightarrow\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{300}\)
31 tháng 3 2019
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 !!!
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.\(\dfrac{5}{12}\)
= \(\dfrac{25}{12}\)
b; \(\dfrac{2}{3}\) : \(\dfrac{1}{2}\) - \(\dfrac{1}{4.8}\)
= \(\dfrac{2}{3}\) x \(\dfrac{2}{1}\) - \(\dfrac{1}{32}\)
= \(\dfrac{4}{3}\) - \(\dfrac{1}{32}\)
= \(\dfrac{125}{96}\)