\(2018\times\frac{3}{4}+2018\times\frac{1}{4}-2018\)
\(=2018\times\left(\frac{3}{4}+\frac{1}{4}-1\right)\)
\(=2018\times0=0\)
B-a:tương tự bài A
b,\(\frac{14}{18}\times\frac{8}{5}-\frac{7}{9}\times\frac{3}{5}\)
\(=\frac{7}{9}\times\frac{8}{5}-\frac{7}{9}\times\frac{3}{5}\)
\(=\frac{7}{9}\times\left(\frac{8}{5}-\frac{3}{5}\right)\)
\(=\frac{7}{9}\times1=\frac{7}{9}\)\
thanks
B-a,\(\frac{3\times125+3\times125}{6\times43+6\times57}=\frac{2\times3\times125}{6\times\left(43+57\right)}\)
\(=\frac{3\times250}{6\times100}=\frac{5}{2\times2}=\frac{5}{4}\)
thanks
A) 2018*3/4+2018*1/4-2018
=2018*(3/4+1/4)-2018
=2018*4/4-2018
=2018*1-2018
=2018-2018
=0
bn thông cảm mình có việc bận,chút nữa mình trả lời câu B