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\(t=\frac{2009.2010+2000}{2011.2010-2020}\)
\(t=\frac{2009.2010+2000}{\left(2009+2\right).2010-2020}\)
\(t=\frac{2009.2010+2000}{2009.2010+2.2010-2020}\)
\(t=\frac{2009.2010+2000}{2009.2010+2000}=1\)
\(q=\frac{\text{2014.2015+2010}}{2016.2015-2020}\)
\(q=\frac{\text{2014.2015+2010}}{\left(2014+2\right).2015-2020}\)
\(q=\frac{\text{2014.2015+2010}}{2014.2015+2.2015-2020}\)
\(q=\frac{\text{2014.2015+2010}}{\text{2014.2015+2010}}=1\)
Ta có:
\(\frac{-2010}{2020}=\frac{-2000+\left(-10\right)}{2010+10}=\frac{-2000}{2010}+\frac{-10}{10}=\frac{-2000}{2010}+1\)
Mà \(\frac{-2000}{2010}+1>\frac{-2000}{2010}\)
\(\Rightarrow\frac{-2010}{2020}>\frac{-2000}{2010}\)
Vậy \(\frac{-2010}{2020}>\frac{-2000}{2010}\).
Ai k mình mình k lại.
\(\frac{-2000}{2010}=0,995...\)
\(\frac{-2010}{2020}=0,995049...\)
vay \(\frac{-2000}{2010}<\frac{-2010}{2020}\)
2009 . 2001 < 2010 .2010 2010 .2007 > 2005. 2009 2011.1998 > 1996.2000 2012. 2000> 2010. 1990 dấu chấm là dấu nhân cho mik k đi ban mik cm
\(\frac{2319}{2010}-\frac{1789}{2000}-\frac{309}{2010}+\frac{2009}{2010}-\frac{211}{2000}\)
\(=\left(\frac{2319}{2010}-\frac{309}{2010}+\frac{2009}{2010}\right)-\left(\frac{1789}{2000}-\frac{211}{2000}\right)\)
\(=\frac{4019}{2010}-\frac{789}{1000}\)
=))
Sai đề rồi.
Đề phải là: \(\frac{1}{1011}+\frac{1}{1012}+\frac{1}{1013}+...+\frac{1}{2020}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\)
Giải như sau:
\(\frac{1}{1011}+\frac{1}{1012}+\frac{1}{1013}+...+\frac{1}{2020}\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2020}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1010}\right)\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2019}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2020}\right)\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\left(đpcm\right).\)
2010.2010 - 2000.2020
= 20102 - (2010 - 10). (2010+10)
= 20102 - (20102 - 102)
= 20102 - 20102 + 102
= 100