(4/15+7/12-19/20)*5/2+1/4

Đáp án:

•$(\frac{4}{15}+\frac{7}{12}-\frac{19}{20}).\frac{5}{2}+\frac{1}{4}$  =0

•$24.(\frac{-1}{5})^{3}+\frac{1}{5}-2.(\frac{-1}{2})^2-\frac{1}{2}$=$\frac{-124}{125}$

Giải thích các bước giải:

  $(\frac{4}{15}+\frac{7}{12}-\frac{19}{20}).\frac{5}{2}+\frac{1}{4}$ 

=$(\frac{4×4+7×5-19×3}{60}).\frac{5}{2}+\frac{1}{4}$ 

=$(\frac{16+35-57}{60}).\frac{5}{2}+\frac{1}{4}$ 

=$\frac{-6}{60}.\frac{5}{2}+\frac{1}{4}$ 

=$\frac{-1}{10}.\frac{5}{2}+\frac{1}{4}$ 

=$\frac{-1}{4}$+$\frac{1}{4}$ 

=0

  $24.(\frac{-1}{5})^{3}+\frac{1}{5}-2.(\frac{-1}{2})^2-\frac{1}{2}$ 

=$24.(\frac{-1}{125})+\frac{1}{5}-2.\frac{1}{4}-\frac{1}{2}$ 

=$\frac{-24}{125}+\frac{1}{5}-\frac{1}{2}-\frac{1}{2}$ 

=$\frac{-24}{125}+\frac{1}{5}-1$ 

=$\frac{-24}{125}+\frac{1}{5}-\frac{5}{5}$ 

=$\frac{-24}{125}-\frac{4}{5}$ 

=$\frac{-24}{125}-\frac{100}{125}$ 

=$\frac{-124}{125}$