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Mình không biết :PPPP
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) \(\left(\frac{4}{3}-\frac{4}{6}\right)+\left(\frac{4}{6}-\frac{4}{9}\right)+\left(\frac{4}{9}-\frac{4}{10}\right)+\left(\frac{4}{12}-\frac{4}{15}\right)\)
\(=\frac{4}{15}-\frac{4}{3}=\frac{-16}{15}\)
C) bạn chỉ ần bỏ các số giống nhau thôi nhé
= 1
b)
=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 = \(\dfrac{502}{3}\)
\(1\cdot\frac{1}{15}\cdot1\frac{1}{16}\cdot1\frac{1}{17}\cdot....\cdot1\frac{1}{2016}\cdot1\frac{1}{2017}\)
\(=\frac{1}{15}\cdot\frac{17}{16}\cdot\frac{18}{17}\cdot....\cdot\frac{2017}{2016}\cdot\frac{2018}{2017}\)
\(=\frac{1}{15}\cdot\frac{1}{16}\cdot2018\)
Dấu "." là dấu nhân nhé bn! phần còn lại bn làm tiếp nha
Đặt biểu thức trên là A ta có:
A = \(\frac{1}{3}\)+ \(\frac{1}{6}\)+ \(\frac{1}{12}\)+ \(\frac{1}{24}\)+ \(\frac{1}{48}\)+ \(\frac{1}{96}\)
A x 3 = \(1\)+ \(\frac{1}{2}\)+ \(\frac{1}{4}\)+ \(\frac{1}{8}\)+ \(\frac{1}{16}\)+ \(\frac{1}{32}\)
A x 3 = \(1\)+ \(1\)- \(\frac{1}{2}\)+ \(\frac{1}{2}\)- \(\frac{1}{4}\)+ \(\frac{1}{4}\)- \(\frac{1}{8}\)+ \(\frac{1}{8}\)- \(\frac{1}{16}\)+ \(\frac{1}{16}\)- \(\frac{1}{32}\)
A x 3 = 2 - \(\frac{1}{32}\)= \(\frac{63}{32}\)
A = \(\frac{63}{32}\): 3 = \(\frac{63}{96}\)
ĐẶT BIỂU THỨC TRÊN LÀ M
TA CÓ \(2M=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+.....+\frac{1}{64}\)
\(\Rightarrow2M-M=1+\frac{1}{2}+\frac{1}{4}+..+\frac{1}{64}-\frac{1}{2}+\frac{1}{4}+..+\frac{1}{128}\)
\(\Rightarrow M=1+\frac{1}{28}\)
A= \(\frac{1}{2}\)+\(\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}\)
2A=2(\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}\))
=1+\(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{64}\)
2A-A= (\(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{64}\)) -(\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}\))
A=1-\(\frac{1}{128}\)
A=\(\frac{127}{128}\)
\(\dfrac{2011\times1990-2011\times999}{2012\times999+1013}\)
\(=\dfrac{2011\times1990-2011}{2012+1013}=\dfrac{2011\times\left(1990-1\right)}{3025}\)
\(=\dfrac{2011\times1989}{3025}=\dfrac{3999879}{3025}=\dfrac{14545}{11}\)