\(a,=\dfrac{5}{7}-\dfrac{33}{8}=-\dfrac{191}{56}\\ b,=\left(\dfrac{12}{17}+\dfrac{5}{17}\right)+\left(\dfrac{19}{7}+\dfrac{3}{7}\right)=1+3=4\\ c,=\left(0,125\cdot8\right)^{12}-\left(\dfrac{45}{15}\right)^3=1-3^3=-26\\ d,=\left(-\dfrac{1}{3}\right)\left(5\dfrac{2}{7}-2\dfrac{2}{7}\right)=-\dfrac{1}{3}\cdot3=-1\\ e,=\dfrac{3^4\cdot3^6}{3^9}=3\)

viết nhầm bài

\(\dfrac{-5}{12}\cdot\dfrac{7}{15}+\dfrac{-5}{12}\cdot\dfrac{8}{15}+\dfrac{5}{12}\)

\(=\dfrac{-5}{12}+\dfrac{5}{12}\)

=0

\(\dfrac{7}{12}>\dfrac{5}{12}\) (Vì tử số 7>5; phân số cùng mẫu số)

\(\dfrac{2}{5}=\dfrac{2.3}{5.3}=\dfrac{6}{15}\) -> Điền dấu "="

\(\dfrac{7}{10}< \dfrac{7}{9}\) (Vì phân số cùng tử, phân số nào có mẫu số lớn hơn thì phân số đó bé hơn. Ta có: 10>9 => 7/10 < 7/9)

\(\dfrac{7}{12}>\dfrac{5}{12}\)

\(\dfrac{2}{5}=\dfrac{6}{15}\)

\(\dfrac{7}{10}< \dfrac{7}{9}\)

7/12.(2/3-5/3+3)

=7/12.2

=7/6

a: =4/5+1/5+2/3+1/3=1+1=2

b: =17/12+7/12+29/7-8/7=3+2=5

c: =3/5+2/5+16/7-1/7-1/7

=1+2=3

d: =2/5+3/5+2/3+1/3+7/4+1/4

=1+1+2

=4

\(\dfrac{4}{5}+\dfrac{2}{3}+\dfrac{1}{5}+\dfrac{1}{3}\)

\(=\left(\dfrac{4}{5}+\dfrac{1}{5}\right)+\left(\dfrac{2}{3}+\dfrac{1}{3}\right)\)

\(=\dfrac{5}{5}+\dfrac{3}{3}\)

\(=1+1\)

\(=2\)

============

\(\dfrac{17}{12}+\dfrac{29}{7}-\dfrac{8}{7}+\dfrac{7}{12}\)

\(=\left(\dfrac{17}{12}+\dfrac{7}{12}\right)+\left(\dfrac{29}{7}-\dfrac{8}{7}\right)\)

\(=\dfrac{24}{12}+\dfrac{21}{7}\)

\(=2+3\)

\(=5\)

====================

\(\dfrac{9}{15}+\dfrac{16}{7}+\dfrac{2}{5}-\dfrac{1}{7}-\dfrac{2}{14}\)

\(=\dfrac{9}{15}+\dfrac{16}{7}+\dfrac{6}{15}-\dfrac{1}{7}-\dfrac{1}{7}\)

\(=\left(\dfrac{9}{15}+\dfrac{6}{15}\right)+\left(\dfrac{16}{7}-\dfrac{1}{7}-\dfrac{1}{7}\right)\)

\(=\dfrac{15}{15}+\dfrac{14}{7}\)

\(=1+2\)

\(=3\)

===============

\(\dfrac{2}{5}+\dfrac{6}{9}+\dfrac{7}{4}+\dfrac{3}{5}+\dfrac{1}{3}+\dfrac{1}{4}\)

\(=\dfrac{2}{5}+\dfrac{2}{3}+\dfrac{7}{4}+\dfrac{3}{5}+\dfrac{1}{3}+\dfrac{1}{4}\)

\(=\left(\dfrac{2}{5}+\dfrac{3}{5}\right)+\left(\dfrac{2}{3}+\dfrac{1}{3}\right)+\left(\dfrac{7}{4}+\dfrac{1}{4}\right)\)

\(=\dfrac{5}{5}+\dfrac{3}{3}+\dfrac{8}{4}\)

\(=1+1+2\)

\(=4\)

\(-\dfrac{7}{2}\)

\(=\dfrac{7}{12}\cdot\left(25\dfrac{3}{5}-31\dfrac{3}{5}\right)=\dfrac{7}{12}\left(-6\right)=-\dfrac{7}{2}\)

g: \(=\dfrac{-3}{4}-\dfrac{1}{4}+\dfrac{5}{7}+\dfrac{2}{7}+\dfrac{3}{5}=\dfrac{3}{5}\)

h: \(=\dfrac{7}{19}\left(\dfrac{8}{11}+\dfrac{3}{11}\right)-\dfrac{12}{19}=\dfrac{7}{19}-\dfrac{12}{19}=-\dfrac{5}{19}\)

i: \(=\dfrac{2013}{7}\left(19+\dfrac{5}{8}-26-\dfrac{5}{8}\right)=\dfrac{2013}{7}\cdot\left(-7\right)=-2013\)

a: \(\dfrac{5}{6}-\dfrac{4}{6}=\dfrac{5-4}{6}=\dfrac{1}{6}\)

b: \(\dfrac{7}{12}-\dfrac{6}{12}=\dfrac{7-6}{12}=\dfrac{1}{12}\)

c: \(\dfrac{7}{9}-\dfrac{2}{9}=\dfrac{7-2}{9}=\dfrac{5}{9}\)

d: \(\dfrac{16}{5}-\dfrac{9}{5}=\dfrac{16-9}{5}=\dfrac{7}{5}\)

a: \(A=\dfrac{7}{12}+\dfrac{5}{72}-\dfrac{11}{36}=\dfrac{42}{72}+\dfrac{5}{72}-\dfrac{22}{72}=\dfrac{25}{72}\)

b: \(B=\dfrac{8+5}{10}:\dfrac{-5}{13}=\dfrac{13}{10}\cdot\dfrac{13}{-5}=-\dfrac{169}{100}\)

c: \(C=\left(\dfrac{88}{132}-\dfrac{33}{132}+\dfrac{60}{132}\right):\left(\dfrac{55}{132}+\dfrac{132}{132}-\dfrac{84}{132}\right)\)

\(=\dfrac{88-33+60}{55+132-84}=\dfrac{115}{103}\)

\(d,-\dfrac{5}{12}+\dfrac{3}{4}=-\dfrac{5}{12}+\dfrac{9}{12}=\dfrac{4}{12}=\dfrac{1}{3}\)

\(e,\dfrac{5}{7}.\dfrac{25}{2}-\dfrac{5}{7}.\dfrac{11}{2}=\dfrac{5}{7}.\left(\dfrac{25}{2}-\dfrac{11}{2}\right)=\dfrac{5}{7}.7=5\)