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a, \(\frac{2}{5}.\frac{1}{3}-\frac{2}{15}:\frac{1}{5}+\frac{3}{5}.\frac{1}{3}\)
\(=\frac{1}{3}.\left(\frac{2}{5}+\frac{3}{5}\right)-\frac{2}{15}.5\)
\(=\frac{1}{3}.1-\frac{2}{3}\)
\(=\frac{1}{3}-\frac{2}{3}\)
\(=\frac{-1}{3}\)
b, \(\left(6-2\frac{4}{5}\right).3\frac{1}{8}+1\frac{3}{8}:\frac{1}{4}\)
\(=\left(6-\frac{14}{5}\right).\frac{25}{8}+\frac{11}{8}.4\)
\(=\frac{16}{5}.\frac{25}{8}+\frac{11}{2}\)
\(=10+\frac{11}{2}\)
\(=\frac{31}{2}\)
1/3×(3/5+2/5)-2/15×1/5
1/3×1-2/15×1/5
1/3-2/15×1/5
1/3-2/75
25/75-2/75
23/75
(6-14/5)×25/8-11/8:4/1
16/5×25/8-11/8:4/1
10/1-11/8:4/1
10/1-11/8×1/4
10/1-11/32
320/32-11/32
309/32
a, 6/7+5/9+8/7-2/9=(6/7+8/7)+(5/9-2/9)=14/7+3/9=2+1/3=7/3
b, 5+7/3=22/3
1-2+3-4+5-6+...+2011-2012
=2012-2011+...+6-5+4-3+2-1
=(2012-2001)+...+(6-5)+(4-3)+(2-1)
=1+1+1+...+1+1(có 1006 số 1)
=1x60
=60
= \(\frac{2}{7}.\left(5\frac{1}{4}-3\frac{1}{4}\right)\)
= \(\frac{2}{7}.2\)
= \(\frac{4}{7}\)
\(\frac{2}{7}.5\frac{1}{4}-\frac{2}{7}.3\frac{1}{4}\)
=> \(\frac{2}{7}.\left(5\frac{1}{4}-3\frac{1}{4}\right)\)
=> \(\frac{2}{7}.2\)
=> \(\frac{4}{7}\)
#Hk_tốt
#Ngọc's_Ken'z