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Ta có:
= (0,6 + 0,415 + 0,005) : 0,01 = 1,02 : 0,01 = 102.
Biểu thức trở thành:
Vậy x = -4.
a) \(\dfrac{-0.8:\left(\dfrac{4}{5}\cdot1.25\right)}{0.64-\dfrac{1}{5}}=\dfrac{\dfrac{-4}{5}:\left(\dfrac{4}{5}\cdot\dfrac{5}{4}\right)}{\dfrac{16}{25}-\dfrac{1}{5}}=\dfrac{\dfrac{-4}{5}:1}{\dfrac{16}{25}-\dfrac{5}{25}}=\dfrac{\dfrac{-4}{5}}{\dfrac{11}{25}}=\dfrac{-4}{5}\cdot\dfrac{25}{11}=\dfrac{-20}{11}\)
b) \(\left(13.71-1\dfrac{5}{6}\right)\cdot6-6\cdot13\cdot17=\left(\dfrac{1371}{100}-\dfrac{11}{6}\right)\cdot6-6\cdot13\cdot17=\dfrac{3563}{300}\cdot6-6\cdot13\cdot17=\dfrac{3563}{50}-6\cdot13\cdot17=\dfrac{3563}{50}-1326=\dfrac{-62737}{50}\)
c) \(\dfrac{\left(\dfrac{3}{5}+0.415+\dfrac{1}{200}\right):0.01}{30.75+\dfrac{1}{12}+3\dfrac{1}{6}}=\dfrac{\left(\dfrac{3}{5}+\dfrac{83}{200}+\dfrac{1}{200}\right):\dfrac{1}{100}}{\dfrac{123}{4}+\dfrac{1}{12}+\dfrac{19}{6}}=\dfrac{\dfrac{51}{50}:\dfrac{1}{100}}{34}=\dfrac{102}{34}=3\)
{\(\frac{112}{200}\)+ 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!!!
\(112\cdot35+112\cdot65+800\)
\(=112\cdot\left(35+65\right)+800\)
\(=112\cdot100+800\)
\(=11200+800\)
\(=12000\)
Ta có: \(M=\dfrac{9}{40}-\dfrac{11}{60}+\dfrac{13}{84}-\dfrac{15}{112}\)
\(=\dfrac{378}{1680}-\dfrac{308}{1680}+\dfrac{260}{1680}-\dfrac{225}{1680}\)
\(=\dfrac{105}{1680}=\dfrac{1}{16}\)
