\(\frac{\left(\frac{11^2}{200}+0,415\right):0,01}{\frac{1}{12}-37,25+3\frac{1}{6}}=\frac{\left(\frac{121}{200}+\frac{83}{200}\right):\frac{1}{100}}{\frac{1}{12}-\frac{149}{4}+\frac{19}{6}}=\frac{\frac{51}{50}.100}{\frac{1}{12}-\frac{447}{12}+\frac{38}{12}}\)

=\(\frac{102}{-34}=-3\)

Bằng -3 nha bạn

AI GIÚP MÌNH ĐIIIII...........!MÌNH CẦN GẤP LẮM

\(-\frac{10}{17}\)

Ta có: \(\frac{\left(\frac{11^2}{200}+0,415\right):0,01}{\frac{1}{12}-37,25+3\frac{1}{6}}=\frac{\left(\frac{121}{200}+\frac{19}{40}\right).100}{\frac{1}{12}-\frac{149}{4}+\frac{17}{6}}\)

\(=\frac{108}{-\frac{103}{3}}=-\frac{324}{103}\)

\(\frac{\frac{75}{100}:\frac{5}{2}+\left(\frac{3}{4}\right)^2-\frac{3}{4}:\frac{149}{4}}{\left(\frac{-121}{200}-\frac{83}{200}\right):\left(\frac{-1}{100}\right)}=\frac{\frac{3}{10}+\frac{9}{16}-\frac{3}{149}}{\frac{-51}{50}:\frac{-1}{100}}=\frac{\frac{69}{80}-\frac{3}{149}}{102}=0,008258487595\)

2) \(\frac{2010+2011}{2011+2012}=\frac{2010}{2011+2012}+\frac{2011}{2011+2012}\)

\(\frac{2010}{2011}>\frac{2010}{2011+2012}\)

\(\frac{2011}{2012}>\frac{2011}{2011+2012}\)

\(\Rightarrow A>B\)

 a) \(1\frac{13}{15}.\left(0,5\right)^2.3+\left(\frac{8}{15}-1\frac{19}{60}\right):1\frac{23}{24}\)

\(=\frac{28}{15}.\frac{1}{4}.3+-\frac{47}{60}:\frac{47}{24}\)

\(=\frac{7}{5}-\frac{2}{5}\)

\(=1\)

b)\(\frac{\left(\frac{11^2}{200}+0,415\right):0,01}{\frac{1}{12}-37,25+3\frac{1}{6}}\)

\(=\frac{\left(\frac{121}{200}+0,415\right):0,01}{\frac{1}{12}-37,25+\frac{19}{6}}\)

\(=\frac{\frac{51}{50}:\frac{1}{100}}{-34}\)

\(=\frac{102}{-34}\)

\(=-3\)

Bài 1:

\(\in\)là thuộc ;     

\(\notin\)là không thuộc ;

\(\subset\)là tập hợp con ;

\(\phi\)là tập hợp rỗng ;

\(\supset\)là 2 tập hợp giao nhau.