A = (112/200 + 0,415) : 0.01 / 1/11 - 37,25 + 3 1/6
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{\(\frac{112}{200}\)+ 0,415} : 0,01
= {\(\frac{112}{200}+\frac{415}{1000}\)} : \(\frac{1}{100}\)
={\(\frac{560}{1000}\)+\(\frac{415}{1000}\)} : \(\frac{1}{100}\)
=\(\frac{975}{1000}\): \(\frac{1}{100}\)
=\(\frac{975}{1000}\)x\(\frac{100}{1}\)
=\(\frac{97500}{1000}\)=\(97,5\)
\(\frac{1}{2}-37,5+3-\frac{1}{6}\)
=\(\frac{1}{2}-\frac{375}{10}+\frac{3}{1}-\frac{1}{6}\)
=\(\frac{15}{30}-\frac{1125}{30}+\frac{90}{30}-\frac{5}{30}\)
=\(\frac{15-1125+90-5}{30}\)
=\(\frac{\left(-1110\right)+85}{30}\)
=\(\frac{-1025}{30}\)
Mik ko chắc!!!
\(\frac{\left(\frac{11^2}{200}+0,415\right):0,01}{\frac{1}{12}-37,25+3\frac{1}{6}}=\frac{\left(\frac{121}{200}+\frac{83}{200}\right):\frac{1}{100}}{\frac{1}{12}-\frac{149}{4}+\frac{19}{6}}=\frac{\frac{51}{50}.100}{\frac{1}{12}-\frac{447}{12}+\frac{38}{12}}\)
=\(\frac{102}{-34}=-3\)
Ta có : \(B=\frac{\left(\frac{11^2}{200}+0,415\right):0,01}{\frac{1}{12}-37,25+3\frac{1}{6}}\)
\(=>B=\frac{\left(\frac{121}{200}+\frac{415}{1000}\right):\frac{1}{100}}{\frac{1}{12}-\frac{3725}{100}+\frac{19}{6}}\)
\(=>B=\frac{\left(\frac{121}{200}+\frac{83}{200}\right)\cdot100}{\frac{1}{12}-\frac{149}{4}+\frac{19}{6}}\)
\(=>B=\frac{\frac{204}{200}\cdot100}{\frac{1}{12}-\frac{447}{12}+\frac{38}{12}}\)
\(=>B=\frac{\frac{204\cdot100}{200}}{-\frac{408}{12}}=\frac{\frac{204}{2}}{-34}=\frac{102}{-34}=-3\)