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a. 1⋅2⋅3+2⋅4⋅6+3⋅6⋅9+4⋅8⋅12
= 6+2⋅4⋅6+3⋅6⋅9+4⋅8⋅12
= 6+48+3⋅6⋅9+4⋅8⋅12
= 6+48+162+4⋅8⋅12
= 6+48+162+384
= 600
b . Ta có \(A=\frac{2010+2011}{2011+2012}=\frac{2010}{2011+2012}+\frac{2011}{2011+2012}.\)
Ta có : \(\frac{2010}{2011+2012}< \frac{2010}{2011}\) và \(\frac{2011}{2011+2012}< \frac{2011}{2012}\)
=> \(\frac{2010+2011}{2011+2012}< \frac{2010}{2011}+\frac{2011}{2012}\)
=> A < B
ta có: 1x3x5x7x9x...x2009x2011
= (1x3x5x7)x(9x11x13x15)x...x2009x2011
= (...5)x(...5)x.....x(....9)
= (....5)x(....9)
= (....5)=> có tận cùng là 5
ủng hộ nhé
neu nhu 1 so nhan voi 1;10;10;1000;...
thi so do cung co so tan cung la 0 nen chu so tan cung cua day :
1*3*5*7...*2009*2011 la 0
\(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{2009.2011}\)=\(\frac{1}{5}-\frac{1}{2011}=\frac{2006}{10055}\)
\(\frac{2006}{10055}+\frac{1}{x}=\frac{1}{5.0,5}\)
=>\(\frac{1}{x}=\frac{1}{5.0,5}-\frac{2006}{10055}\)
\(\frac{1}{x}=\frac{1}{2,5}-\frac{2006}{10055}\)
\(\frac{1}{x}=\frac{4022}{10055}\)
x=1 x 10055 : 4022
x = 2,5
--------------------Hết-------------------
( 35 x 18 - 9 x 70 ) x 2010 x 2011 x ... x 2020
= (35 x 18 - 9 x 35 x 2) x 2010 x 2011 x ... x 2020
= ( 35 x 18 - 18 x 35 ) x 2010 x 2011 x ... x 2020
= 35 x (18 - 18 ) x 2010 x 2011 x ... x 2020
= 35 x 0 x 2010 x 2011 x ... x 2020
= 0 x 2010 x 2011 x ... x 2020
= 0
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{x.\left(x+2\right)}=\frac{20}{41}\)
\(\Rightarrow\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{x.\left(x+2\right)}\right)=\frac{20}{41}\)
\(\Rightarrow\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{20}{41}\)
\(\Rightarrow\frac{1}{2}.\left(1-\frac{1}{x+2}\right)=\frac{20}{41}\)
\(\Rightarrow1-\frac{1}{x+2}=\frac{20}{41}:\frac{1}{2}\)
\(\Rightarrow1-\frac{1}{x+2}=\frac{40}{41}\)
\(\Rightarrow\frac{1}{x+2}=1-\frac{40}{41}=\frac{1}{41}\)
=> x + 2 = 41
=> x = 39