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a)ta có : 2017/2018 = (2018 - 1) / 2018 = 2018/2018 - 1/2018 = 1 - 1/2018
Lại có : 9/10 = (10-1)/10 = 10/10-1/10 = 1-1/10
Vì 2018>10 => 1/2018 < 1/10
=> 1-1/2018 > 1-1/10
=> 2017/2018 > 9/10
Vậy 2017/2018 > 9/10
b) ta có : 8/5 = (5+3)/5 = 5/5 + 3/5 = 1 + 3/5
lại có : 2017/2014 = (2014+3)/2014 = 2014/2014 + 3/2014 = 1 + 3/2014
vì 5<2014 => 3/5 > 3/2014 => 1+ 3/5 > 1+ 3/2014
=> 8/5 > 2017/2014
vậy...
đó .bạn dựa vào đó làm mấy câu sau nha.Chúc bạn học giỏi.nếu bạn cần thì mk sẽ giải hết.
Vì \(\frac{1}{33}>\frac{1}{34}>\frac{1}{35}>\frac{1}{36}\)
\(\Rightarrow M>\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}\)\(\)
\(\Rightarrow M>\frac{4}{36}=\frac{1}{9}\)
Mà \(\frac{1}{9}>\frac{1}{10}\)
\(\Rightarrow\)\(M>\frac{1}{9}>\frac{1}{10}\)
Vậy : M > N
a) Ta thấy \(\frac{1}{2}< \frac{2}{3};\frac{3}{4}< \frac{4}{5};...;\frac{99}{100}< \frac{100}{101}\)
\(\Rightarrow A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}< B=\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}\)
b) \(A.B=\left(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}\right).\left(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}\right)\)
\(A.B=\frac{1.\left(3.5...99\right).\left(2.4.6...100\right)}{\left(2.4.6...100\right).\left(3.5.7...99\right).101}=\frac{1}{101}\)
c) vì A < b nên A . A < A . B < \(\frac{1}{101}< \frac{1}{100}\)
do đó : A . A < \(\frac{1}{10}.\frac{1}{10}\)suy ra A < \(\frac{1}{10}\)
= -1 + -1 + -1 + -1 +...+ -1 + -1
dãy trên có số số hạng là :
(2018- 1):1 + 1 = 2016
vậy có 1008 số 1
= -1008
tk nha, bài này mk làm rồi
#)Giải :
\(\left(\frac{2012}{2015}+\frac{2011}{2016}+\frac{2010}{2016}+\frac{2009}{2018}\right)\left(\frac{1}{6}+\frac{1}{3}+\frac{1}{2}\right)\)
\(=\left(\frac{2012}{2015}+\frac{2011}{2016}+\frac{2010}{2016}+\frac{2009}{2018}\right)\left(\frac{1}{2}+\frac{1}{2}\right)\)
\(=\left(\frac{2012}{2015}+\frac{2011}{2016}+\frac{2010}{2016}+\frac{2009}{2018}\right)\times0\)
\(=0\)
\(\left(\frac{2012}{2015}+\frac{2011}{2016}+\frac{2010}{2017}+\frac{2009}{2018}\right).\left(\frac{1}{6}+\frac{1}{3}+\frac{1}{2}\right)\)
\(=\left(\frac{2012}{2015}+\frac{2011}{2016}+\frac{2010}{2017}+\frac{2009}{2018}\right).\left(\frac{1}{6}+\frac{2}{6}+\frac{3}{6}\right)\)
=\(\left(\frac{2012}{2015}+\frac{2011}{2016}+\frac{2010}{2017}+\frac{2009}{2018}\right).0\)
\(=0\)
1.
a) \(\frac{6}{15}+\frac{6}{35}+\frac{6}{63}+\frac{6}{99}+\frac{6}{143}\)
\(=\frac{6}{3.5}+\frac{6}{5.7}+\frac{6}{7.9}+\frac{6}{9.11}+\frac{6}{11.13}\)
\(=\frac{6}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{11}-\frac{1}{13}\right)\)
\(=\frac{6}{2}\left(\frac{1}{3}-\frac{1}{13}\right)\)
\(=\frac{6}{2}.\frac{10}{39}\)
\(=\frac{10}{13}\)
b) \(\frac{3}{24}+\frac{3}{48}+\frac{3}{80}+\frac{3}{120}+\frac{3}{168}\)
\(=\frac{3}{4.6}+\frac{3}{6.8}+\frac{3}{8.10}+\frac{3}{10.12}+\frac{3}{12.14}\)
\(=\frac{3}{2}\left(\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+...+\frac{1}{12}-\frac{1}{14}\right)\)
\(=\frac{3}{2}.\left(\frac{1}{4}-\frac{1}{14}\right)\)
\(=\frac{3}{2}.\frac{5}{28}\)
\(=\frac{15}{56}\)
\(a.\frac{6}{3.5}+\frac{6}{5.7}+...+\frac{6}{11.13}\)
\(=3.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)\)
\(=3.\left(\frac{1}{3}-\frac{1}{13}\right)\)
\(=3.\frac{10}{39}\)
\(=\frac{10}{13}\)