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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
\(\frac{22}{7}\)> \(\frac{11}{5}\)vì 22 : 7 = 3,14 ; 11: 5 = 2,2
\(\frac{15}{59}\)< \(\frac{24}{97}\)vì 15 : 59 = 0,21 ; 24 : 97 = 0,24
\(\frac{11}{19}\)< \(\frac{13}{18}\)vì 11 : 19 = 0,57 ; 13 : 18 = 0,72
\(\frac{7}{10}\)> \(\frac{4}{9}\)vì 7 : 10 = 0,7 ; 4 : 9 = 0,44
A=1011-1/1012-1<1
=>A=1011-1/1012-1<1011+1.10/1012+1.10
=1011+10/1012+10
=10(1010+1)/10(1011+1)
=1010+1/1011+1=B
=>A<B
Ta có : \(\frac{11}{12}=\frac{11.10}{12.10}=\frac{110}{120}\)
\(\frac{9}{10}=\frac{9.12}{10.12}=\frac{108}{120}\)
Ta thấy \(\frac{110}{120}>\frac{108}{120}\Rightarrow\frac{11}{20}>\frac{9}{10}\)
Vậy \(\frac{11}{20}>\frac{9}{10}\)
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
a. 19/10 > 10/11
b. 11/10 = 12/11
c. 9/10 = 10/11
a)\(\dfrac{19}{10}>\dfrac{10}{11}\)
b)\(\dfrac{11}{10}=\dfrac{12}{11}\)
c)\(\dfrac{9}{10}< \dfrac{10}{11}\)