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a; (5142 - 17 x 8 + 242 : 11) x (27 - 3 x 9)
= (5142 - 17 x 8 + 242 : 11) x (27 - 27)
= (5142 - 17 x 8 + 242 : 11) x 0
= 0
b;
(1 + \(\dfrac{1}{2}\)) \(\times\) (1 + \(\dfrac{1}{3}\)) \(\times\) ( 1 + \(\dfrac{1}{4}\)) \(\times\) ... \(\times\) (1 + \(\dfrac{1}{2010}\)) \(\times\)(1 + \(\dfrac{1}{2011}\))
= \(\dfrac{2+1}{2}\) \(\times\) \(\dfrac{3+1}{3}\) \(\times\) \(\dfrac{4+1}{4}\)\(\times\) ... \(\times\) \(\dfrac{2010+1}{2010}\)\(\times\) \(\dfrac{2011+1}{2011}\)
= \(\dfrac{3}{2}\)\(\times\)\(\dfrac{4}{3}\)\(\times\)\(\dfrac{5}{4}\)\(\times\)...\(\times\)\(\dfrac{2011}{2010}\)\(\times\)\(\dfrac{2012}{2011}\)
= \(\dfrac{2012}{2}\)
= 1006
=13/12x14/13x15/14x16/15x...x2006/2005x2007/2006x2008/2007
=2008/12
=502/3
A = 1\(\dfrac{1}{12}\) \(\times\) 1\(\dfrac{1}{13}\) \(\times\) 1\(\dfrac{1}{14}\) \(\times\) 1\(\dfrac{1}{15}\) \(\times\) ... \(\times\) 1\(\dfrac{1}{2005}\) \(\times\) 1\(\dfrac{1}{2006}\) \(\times\) 1\(\dfrac{1}{2007}\)
A = ( 1 + \(\dfrac{1}{12}\)) \(\times\) ( 1 + \(\dfrac{1}{13}\)) \(\times\) ( 1 + \(\dfrac{1}{14}\)) \(\times\)...\(\times\) ( 1 + \(\dfrac{1}{2006}\))\(\times\)(1+\(\dfrac{1}{2007}\))
A = \(\dfrac{13}{12}\) \(\times\) \(\dfrac{14}{13}\) \(\times\) \(\dfrac{15}{14}\) \(\times\) ...\(\times\) \(\dfrac{2007}{2006}\) \(\times\) \(\dfrac{2008}{2007}\)
A = \(\dfrac{13\times14\times15\times...\times2007}{13\times14\times15\times...\times2007}\) \(\times\) \(\dfrac{2008}{12}\)
A = 1 \(\times\) \(\dfrac{502}{3}\)
A = \(\dfrac{502}{3}\)
Ta có công thức tổng quát:
\(\dfrac{k}{n\cdot\left(n+k\right)}=\dfrac{1}{n}-\dfrac{1}{n+k}\)
\(a,A=\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+...+\dfrac{1}{x\left(x+3\right)}\\ =\dfrac{1}{3}\left(\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+...+\dfrac{3}{x\left(x+3\right)}\right)\\ =\dfrac{1}{3}\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)\\ =\dfrac{1}{3}\cdot\left(\dfrac{1}{5}-\dfrac{1}{x+3}\right)\\ =\dfrac{1}{3}\cdot\dfrac{x-2}{5\left(x+3\right)}\\ =\dfrac{x-2}{15\left(x+3\right)}\)
Theo đề bài ta có:
\(A=\dfrac{101}{1540}\\ \Rightarrow\dfrac{x-2}{15\left(x+3\right)}=\dfrac{101}{1540}\\ \Rightarrow\dfrac{x-2}{x+3}=\dfrac{303}{308}\\ \Rightarrow\dfrac{x-2}{x+3}=\dfrac{305-2}{305+3}\\ \Rightarrow x=305\)
a; A = \(\dfrac{4026\times2014+4030}{2013\times2016-2011}\)
A = \(\dfrac{2\times\left(2013\times2014+2015\right)}{2013\times2016-2011}\)
A = \(\dfrac{2\times\left(2013\times2016-2013\times2+2015\right)}{2013\times2016-2011}\)
A = \(\dfrac{2\times\left(2013\times2016-4026+2015\right)}{2013\times2016-2011}\)
A = \(\dfrac{2\times\left(2013\times2016-2011\right)}{2013\times2016-2011}\)
A = 2
a) \(\dfrac{2}{3}+\dfrac{3}{5}=\dfrac{10}{15}+\dfrac{9}{15}=\dfrac{19}{15}\)
a) \(\dfrac{7}{12}-\dfrac{2}{7}+\dfrac{1}{12}=\dfrac{2}{3}-\dfrac{2}{7}=\dfrac{14}{21}-\dfrac{6}{21}=\dfrac{8}{21}\)
\(71+52,5\times4=\frac{x+140}{x}+210\)
\(71+210=\frac{x+140}{x}+210\)
\(=>\frac{x+140}{x}=71\)
\(71=\frac{142}{2}\)\(\Rightarrow x=142-140=2\)
Sau khi lấy ra lần 1, trong kho gạo còn lại số phần gạo là:
\(1-\dfrac{1}{13}=\dfrac{12}{13}\) (phần số gạo)
Số gạo lần 2 lấy ra chiếm số phần so với số gạo ban đầu là:
\(\dfrac{1}{11}\times\dfrac{12}{13}=\dfrac{12}{143}\) (phần số gạo)
Sau khi lấy ra lần 2 trong kho còn lại là:
\(\dfrac{12}{13}-\dfrac{12}{143}=\dfrac{120}{143}\) (phần số gạo)
Số gạo trong kho ban đầu là:
\(5040:\dfrac{120}{143}=6006\) (kg)
Phân số chỉ 5040 kg gạo là:
1 - \(\dfrac{1}{11}\) = \(\dfrac{10}{11}\) (số gạo còn lại sau lần lấy thứ nhất)
Số gạo còn lại sau lần lấy thứ nhất là:
5040 : \(\dfrac{10}{11}\) = 5544 (kg)
Phân số chỉ 5544 kg gạo là:
1 - \(\dfrac{1}{13}\) = \(\dfrac{12}{13}\) (số gạo trong kho)
Số gạo trong kho là:
5544 : \(\dfrac{12}{13}\) = 6006 (kg)
Đáp số:..
( 1 + \(\dfrac{2}{5}\)).( 1 + \(\dfrac{2}{7}\) ).( 1 + \(\dfrac{2}{9}\)).( 1 + \(\dfrac{2}{11}\)).....( 1 + \(\dfrac{2}{97}\)).( 1 + \(\dfrac{2}{99}\))
= \(\dfrac{7}{5}\).\(\dfrac{9}{7}\).\(\dfrac{11}{9}\).\(\dfrac{13}{11}\)...\(\dfrac{99}{97}\).\(\dfrac{101}{99}\)
= \(\dfrac{7.9.11.13...99.101}{5.7.9.11...97.99}\)
= \(\dfrac{101}{5}\)
❤ HOK TT ❤