Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) Xét \(\frac{1999.2000-2}{1998.1999+3997}=\frac{1999.\left(1998+2\right)-2}{1998.1999+3997}=\frac{1999.1998+1999.2-2}{1998.1999+3997}=\frac{1999.1998+3996}{1998.1998+3997}\)
=> A < B
A=54.107-53/53.107+54
=(53+1).107-53/53.107+54
=53*107+107-53/53.107+54
=53.107+54/53.107+54
=1
B=135.269-133/134.269+135
=(134+1).269-133/134.269+135
=134.296+296-133/134.269+135
=134.269+136/134.269+135
Vì 134.269+136/134.269+135>1 . Nên B>A
A)\(-\frac{3}{29}+\frac{16}{58}=-\frac{3}{29}+\frac{8}{29}=\frac{5}{29}\)
B)\(-\frac{1}{5}+\frac{5}{18}+-\frac{7}{9}=-\frac{1}{5}+\frac{5}{18}+-\frac{14}{18}=-\frac{1}{5}+-\frac{1}{2}\)\(=-\frac{2}{10}+-\frac{5}{10}=-\frac{7}{10}\)
C)\(-\frac{8}{18}+-\frac{15}{27}=-\frac{4}{9}+-\frac{5}{9}=-1\)
b) D = \(\frac{3}{4}+\frac{3}{8}+\frac{3}{70}+\frac{3}{130}+\frac{3}{208}+\frac{3}{304}\)
D = \(3\left(\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+\frac{1}{130}+\frac{1}{208}+\frac{1}{304}\right)\)
D = \(3\left(\frac{1}{1x4}+\frac{1}{4x7}+\frac{1}{7x10}+\frac{1}{10x13}+\frac{1}{13x16}+\frac{1}{16x19}\right)\)
D = \(\frac{1}{1}-\frac{1}{19}=\frac{18}{19}\)
Chắc vậy
Vậy thì chọn đúng đi chứ !
\(A=\frac{54.107-53}{53.107+54}=\frac{53.107+107-53}{53.107+54}=\frac{53.107+54}{53.107+54}=1\)
\(B=\frac{135.269-133}{134.269+135}=\frac{134.269+269-133}{134.269+135}=\frac{134.269+136}{134.269+135}>1\)
Vậy A > B