c: \(=\dfrac{4}{9}\left(\dfrac{5}{7}+\dfrac{2}{5}+\dfrac{2}{7}-\dfrac{7}{5}\right)=\dfrac{4}{9}\cdot\left(1-1\right)=0\)

d: \(=\dfrac{4-3-1}{12}\cdot\left(\dfrac{67}{111}+\dfrac{2}{33}-\dfrac{15}{117}\right)=0\cdot\left(\dfrac{67}{111}+\dfrac{2}{33}-\dfrac{15}{117}\right)=0\)

\(\left(-\dfrac{1}{2}\right)^2\div\dfrac{1}{4}-2\times\left(-\dfrac{1}{2}\right)^2\\= \dfrac{1}{4}\div\dfrac{1}{4}-2\times\dfrac{1}{4}\\ =1-\dfrac{1}{2}\\ =\dfrac{1}{2}\)

\(\left(-2\right)^3\times-\dfrac{1}{24}+\left(\dfrac{4}{3}-1\dfrac{5}{6}\right)\div\dfrac{5}{12}\)

=  \(-6\times-\dfrac{1}{24}+\left(\dfrac{4}{3}-\dfrac{11}{6}\right)\div\dfrac{5}{12}\)

=  \(\dfrac{1}{4}+-\dfrac{1}{2}\div\dfrac{5}{12}\)

=  \(\dfrac{1}{4}+-\dfrac{6}{5}\)

=  \(\dfrac{1}{4}-\dfrac{6}{5}\)

=  \(-\dfrac{19}{20}\)

\(\left(6\dfrac{4}{9}+\dfrac{7}{11}\right)-\left(4\dfrac{4}{9}-2\dfrac{4}{11}\right)\\ =\dfrac{58}{9}+\dfrac{7}{11}-\dfrac{40}{9}+\dfrac{26}{11}\\ =\dfrac{58}{9}-\dfrac{40}{9}+\dfrac{7}{11}+\dfrac{26}{11}\\ =12+3\\ =15\)

\(a,\left(\dfrac{-1}{2}\right)^2:\dfrac{1}{4}-2\left(-\dfrac{1}{2}\right)^2\)

\(=\left(-\dfrac{1}{2}\right)^2\left(4-2\right)\)

\(=\dfrac{1}{4}.2=\dfrac{1}{2}\)

\(b,\left(-2\right)^3.\dfrac{-1}{24}+\left(\dfrac{4}{3}-1\dfrac{5}{6}\right):\dfrac{5}{12}\)

\(=\left(-8\right).\dfrac{-1}{24}+\left(-\dfrac{1}{2}\right).\dfrac{12}{5}\)

\(=\dfrac{1}{3}+\left(-\dfrac{1}{5}\right)=\dfrac{2}{15}\)

\(c,\left(6\dfrac{4}{9}+\dfrac{7}{11}\right)-\left(4\dfrac{4}{9}-2\dfrac{4}{11}\right)\)

\(=\dfrac{701}{99}-\dfrac{206}{99}=\dfrac{495}{99}=5\)

\(d,10\dfrac{1}{5}-5\dfrac{1}{2}.\dfrac{60}{11}+\dfrac{3}{15\%}\)

\(=\dfrac{51}{5}-30+20=\dfrac{1}{5}\)

\(e,\dfrac{5}{7}.\dfrac{5}{11}+\dfrac{5}{7}.\dfrac{2}{11}-\dfrac{5}{7}.\dfrac{14}{11}\)

\(=\dfrac{5}{7}\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)=\dfrac{5}{7}.\left(-\dfrac{7}{11}\right)\)

\(=-\dfrac{5}{11}\)

\(f,\dfrac{-5}{7}.\dfrac{2}{11}+\left(-\dfrac{5}{7}\right).\dfrac{9}{11}+1\dfrac{5}{7}\)

\(=\left(-\dfrac{5}{7}\right)\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+\dfrac{12}{7}\)

\(=\left(-\dfrac{5}{7}\right)+\dfrac{12}{7}=1\)

\(10\dfrac{1}{5}-5\dfrac{1}{2}\times\dfrac{60}{11}+3\div15\%\\ =\dfrac{51}{5}-\dfrac{11}{2}\times\dfrac{60}{11}+3\div\dfrac{15}{100}\\ =\dfrac{51}{5}-30+20\\ =10,2-30+20\\ =0,2\)

\(\dfrac{5}{7}\times\dfrac{5}{11}+\dfrac{5}{7}\times\dfrac{2}{11}-\dfrac{5}{7}\times\dfrac{14}{11}\\ =\dfrac{5}{7}\times\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)\\ =\dfrac{5}{7}\times-\dfrac{7}{11}=-\dfrac{5}{11}\)

\(-\dfrac{5}{7}\times\dfrac{2}{11}+-\dfrac{5}{7}\times\dfrac{9}{11}+1\dfrac{5}{7}\\ =-\dfrac{5}{7}\times\dfrac{2}{11}+-\dfrac{5}{7}\times\dfrac{9}{11}+\dfrac{12}{7}\\ =-\dfrac{5}{7}\times\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+\dfrac{12}{7}\\ =-\dfrac{5}{7}\times1+\dfrac{12}{7}\\ =1\)

a/\(\frac{\left(2^3.5.7\right).\left(5^2.7^3\right)}{\left(2.5.7^2\right)^2}\)

=\(\frac{2^3.5^3.7^4}{2^2.5^2.7^4}\)

=2.5

=10

nhầm:mux3=mũ3

là số nguyên tố 

1) \(5\frac{2}{7}\).\(\frac{8}{11}\)+\(5\frac{2}{7}.\frac{5}{11}+5\frac{2}{7}.\frac{-2}{11}\)

= \(5\frac{2}{7}\).[\(\frac{8}{11}+\frac{5}{11}+\frac{-2}{11}\)]

= \(5\frac{2}{7}\).1

=\(5\frac{2}{7}\)

2)

5 2/7.8/9+5 2/7.5/11-5 2/7.2/11

=5 2/7.(8/11+5/11+2/11)

=5 2/7.15/11

=240/77.

(67/111+2/33-15/117).(1/3-1/4-1/12)

=811/1221.1/12

=811/14652

Số số hạng của dãy:(2010-7):1+1=2004(số)

Vậy có tất cả:2004:2=1002(cặp)

A=7-8+9-10+11-12+...+2009-2010

A=(7-8)+(9-10)+(11-12)+...+(2009-2010)

A=-1+(-1)+(-1)+...+(-1)

Vậy A=(-1)*1002=-1002