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\(\frac{\left(\frac{11^2}{200}+0,415\right):0,01}{\frac{1}{12}-37,25+3\frac{1}{6}}=\frac{\left(\frac{121}{200}+\frac{83}{200}\right):\frac{1}{100}}{\frac{1}{12}-\frac{149}{4}+\frac{19}{6}}=\frac{\frac{51}{50}.100}{\frac{1}{12}-\frac{447}{12}+\frac{38}{12}}\)
=\(\frac{102}{-34}=-3\)
\(\frac{\frac{75}{100}:\frac{5}{2}+\left(\frac{3}{4}\right)^2-\frac{3}{4}:\frac{149}{4}}{\left(\frac{-121}{200}-\frac{83}{200}\right):\left(\frac{-1}{100}\right)}=\frac{\frac{3}{10}+\frac{9}{16}-\frac{3}{149}}{\frac{-51}{50}:\frac{-1}{100}}=\frac{\frac{69}{80}-\frac{3}{149}}{102}=0,008258487595\)
2) \(\frac{2010+2011}{2011+2012}=\frac{2010}{2011+2012}+\frac{2011}{2011+2012}\)
\(\frac{2010}{2011}>\frac{2010}{2011+2012}\)
\(\frac{2011}{2012}>\frac{2011}{2011+2012}\)
\(\Rightarrow A>B\)
- Bài 3:
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)\(\frac{\left(\frac{11^2}{200}+0,415\right):0,01}{\frac{1}{12}-37,25+3\frac{1}{6}}\)
\(=\frac{\left(\frac{121}{200}+0,415\right):0,01}{\frac{1}{12}-37,25+\frac{19}{6}}\)
\(=\frac{\frac{51}{50}:\frac{1}{100}}{-34}\)
\(=\frac{102}{-34}\)
\(=-3\)
a, \(1\dfrac{13}{15}.\left(0,5\right)^2.3+\left(\dfrac{8}{15}-1\dfrac{19}{60}\right):1\dfrac{23}{24}\)
= \(\dfrac{28}{15}.\dfrac{25}{100}.3+\left(\dfrac{8}{15}-\dfrac{79}{60}\right):\dfrac{47}{24}\)
= \(\dfrac{28}{15}.\dfrac{1}{4}.3+\left(\dfrac{32-79}{60}\right).\dfrac{24}{47}\)
= \(\dfrac{84}{60}+\dfrac{-47}{60}.\dfrac{24}{47}\)
= \(\dfrac{84}{60}+\dfrac{-24}{60}=\dfrac{60}{60}=1\)
b, \(\dfrac{\left(\dfrac{11^2}{200}+0,415\right):0,01}{\dfrac{1}{12}-37,25+3\dfrac{1}{6}}\)
= \(\dfrac{\left(\dfrac{121}{200}+\dfrac{415}{1000}\right):\dfrac{1}{100}}{\dfrac{1}{12}-\dfrac{3725}{100}+\dfrac{19}{6}}=\dfrac{\left(\dfrac{121}{200}+\dfrac{83}{200}\right).100}{\dfrac{1}{12}-\dfrac{149}{4}+\dfrac{19}{6}}\)
= \(\dfrac{\dfrac{51}{50}.100}{-34}=\dfrac{102}{-34}=-3\)
Bằng -3 nha bạn