a) Ta có: \(\left(\dfrac{9}{25}-2\cdot18\right):\left(3\dfrac{4}{5}+0.2\right)\)

\(=\left(\dfrac{9}{25}-36\right):\left(\dfrac{19}{5}+\dfrac{1}{5}\right)\)

\(=\left(\dfrac{9}{25}-\dfrac{900}{25}\right):\dfrac{20}{5}\)

\(=\dfrac{-891}{25}\cdot\dfrac{1}{4}\)

\(=-\dfrac{891}{100}\)

b) Ta có: \(\dfrac{3}{8}\cdot19\dfrac{1}{3}+\dfrac{3}{8}\cdot33\dfrac{1}{3}\)

\(=\dfrac{3}{8}\cdot\dfrac{58}{3}+\dfrac{3}{8}\cdot\dfrac{100}{3}\)

\(=\dfrac{58}{8}+\dfrac{100}{8}\)

\(=\dfrac{158}{8}=\dfrac{79}{4}\)

c) Ta có: \(15\cdot\left(-\dfrac{2}{3}\right)^2-\dfrac{7}{3}\)

\(=15\cdot\dfrac{4}{9}-\dfrac{7}{3}\)

\(=\dfrac{20}{3}-\dfrac{7}{3}\)

\(=\dfrac{13}{3}\)

d) Ta có: \(\dfrac{1}{2}\sqrt{64}-\sqrt{\dfrac{4}{25}}+\left(-1\right)^{2007}\)

\(=\dfrac{1}{2}\cdot8-\dfrac{2}{5}-1\)

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

\(=3-\dfrac{2}{5}\)

\(=\dfrac{15}{5}-\dfrac{2}{5}=\dfrac{13}{5}\)

e) Ta có: \(\left(-\dfrac{5}{2}\right)^2:\left(-15\right)-\left(0.45+\dfrac{3}{4}\right)\cdot\left(-1\dfrac{5}{9}\right)\)

\(=\dfrac{25}{4}\cdot\dfrac{-1}{15}-\left(\dfrac{9}{20}+\dfrac{15}{20}\right)\cdot\dfrac{-14}{9}\)

\(=\dfrac{-25}{60}-\dfrac{24}{20}\cdot\dfrac{-14}{9}\)

\(=\dfrac{-25}{60}+\dfrac{28}{15}\)

\(=\dfrac{-25}{60}+\dfrac{112}{60}\)

\(=\dfrac{87}{60}=\dfrac{29}{20}\)

f) Ta có: \(\left(-\dfrac{1}{3}\right)-\left(-\dfrac{3}{5}\right)^0+\left(1-\dfrac{1}{2}\right)^2:2\)

\(=-\dfrac{1}{3}-1+\left(\dfrac{1}{2}\right)^2\cdot\dfrac{1}{2}\)

\(=\dfrac{-4}{3}+\dfrac{1}{4}\cdot\dfrac{1}{2}\)

\(=\dfrac{-4}{3}+\dfrac{1}{8}\)

\(=\dfrac{-32}{24}+\dfrac{3}{24}=\dfrac{-29}{24}\)

g) Ta có: \(\left(\dfrac{1}{2}\right)^{15}\cdot\left(\dfrac{1}{4}\right)^{20}\)

\(=\left(\dfrac{1}{2}\right)^{15}\cdot\left(\dfrac{1}{2}\right)^{40}\)

\(=\left(\dfrac{1}{2}\right)^{55}\)

\(=\dfrac{1}{2^{55}}\)

h) Ta có: \(\dfrac{5^4\cdot20}{25^5\cdot4^5}\)

\(=\dfrac{5^4\cdot5\cdot2^2}{5^{10}\cdot2^{10}}\)

\(=\dfrac{5^5}{5^{10}}\cdot\dfrac{2^2}{2^{10}}\)

\(=\dfrac{1}{5^5}\cdot\dfrac{1}{2^8}\)

\(=\dfrac{1}{800000}\)

Ta có\(\dfrac{3}{8}.19\dfrac{1}{3}-\dfrac{3}{8}.33\dfrac{1}{3}=\dfrac{3}{8}\left(19\dfrac{1}{3}-33\dfrac{1}{3}\right)=\dfrac{3}{8}.\left(-14\right)=-\dfrac{21}{4}=-5,25\)

\(\left(\dfrac{9}{25}-2.18\right):\left(3\dfrac{4}{5}+0,2\right)\)

=\(\left(\dfrac{9}{25}-36\right):4\)

= -35,64:4=-8,91

a)Ta có:\(\left(\dfrac{9}{25}-2,18\right):\left(3\dfrac{4}{5}+0,2\right)=-1,82:4=-\dfrac{91}{200}=-0,455\)

b)Ta có:\(\dfrac{3}{8}.19\dfrac{1}{3}-\dfrac{3}{8}.33\dfrac{1}{3}=\dfrac{3}{8}\left(19\dfrac{1}{3}-33\dfrac{1}{3}\right)=\dfrac{3}{8}.\left(-14\right)=-\dfrac{21}{4}=-5,25\)

c)Ta có:\(1\dfrac{4}{23}+\dfrac{5}{21}-\dfrac{4}{23}+0,5+\dfrac{16}{21}=(1\dfrac{4}{23}-\dfrac{4}{23})+\left(\dfrac{5}{21}+\dfrac{16}{21}\right)+0,5=1+1+0,5=2,5\)Chúc bạn học tốt!!

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

Gợi ý: Sử dụng tính chất phân phối của phép nhân đối với phép cộng để nhóm thừa số chung ra ngoài.

a) \(\dfrac{3}{5}+\dfrac{7}{25}=\dfrac{15}{25}+\dfrac{7}{25}=\dfrac{15+7}{25}=\dfrac{22}{25}\)

b) \(\dfrac{8}{11}-\dfrac{19}{33}=\dfrac{24}{33}-\dfrac{19}{33}=\dfrac{24-19}{33}=\dfrac{5}{33}\)

c) \(\dfrac{16}{21}\times\dfrac{3}{5}=\dfrac{16\times3}{21\times5}=\dfrac{48}{105}=\dfrac{16}{35}\)

d) \(\dfrac{14}{41}\div\dfrac{7}{9}=\dfrac{14}{41}\times\dfrac{9}{7}=\dfrac{14\times9}{41\times7}=\dfrac{126}{287}=\dfrac{18}{41}\)