\(\dfrac{5}{3}\cdot\dfrac{7}{25}+\dfrac{5}{3}\cdot\dfrac{21}{25}-\dfrac{5}{3}\cdot\dfrac{7}{25}\)

\(=\dfrac{5}{3}\cdot\left(\dfrac{7.}{25}+\dfrac{21}{25}-\dfrac{7}{25}\right)\)

\(=\dfrac{5}{3}\cdot\dfrac{21}{25}=\dfrac{7}{5}\)

b) \(250\%+19\dfrac{3}{11}\cdot\dfrac{7}{26}-6\dfrac{3}{11}\cdot\dfrac{7}{26}\)

\(=\dfrac{5}{2}+\dfrac{212}{11}\cdot\dfrac{7}{26}-\dfrac{69}{11}\cdot\dfrac{7}{26}\)

\(=\dfrac{7}{26}\cdot\left(\dfrac{212}{11}-\dfrac{69}{11}\right)+\dfrac{5}{2}\)

\(=\dfrac{7}{26}\cdot13+\dfrac{5}{2}\)

\(=\dfrac{7}{2}+\dfrac{5}{2}\)

\(=\dfrac{12}{2}=6\)

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

Bài 1:

\(a,=\frac{2}{3}-\frac{16}{3}=\frac{-14}{3}\)

\(b,=\left(\frac{3}{7}+\frac{4}{7}\right)+\left(-\frac{6}{19}+\frac{-13}{19}\right)=1-1=0\)

\(c,=\frac{3}{5}.\left(\frac{8}{9}-\frac{7}{9}+\frac{26}{9}\right)=\frac{3}{5}.3=\frac{9}{5}\)

a,\(\dfrac{1}{2}\).\(\dfrac{4}{3}\)-\(\dfrac{20}{3}\).\(\dfrac{4}{5}\)=\(\dfrac{2}{3}\)-\(\dfrac{16}{3}\)=-\(\dfrac{14}{3}\)

giúp mk vs mn ơi , mai cô giáo ktra mk r

a: \(=\dfrac{8}{9}\cdot\dfrac{9}{4}\cdot\dfrac{12}{19}\cdot\dfrac{19}{24}=\dfrac{1}{2}\cdot2=1\)

b: \(=\dfrac{5}{16}\cdot\dfrac{17}{15}\cdot\dfrac{8}{17}=\dfrac{5}{16}\cdot\dfrac{8}{15}=\dfrac{40}{240}=\dfrac{1}{6}\)

c: \(=\dfrac{4}{13}\left(\dfrac{2}{7}+\dfrac{5}{7}\right)-\dfrac{3}{26}=\dfrac{4}{13}-\dfrac{3}{26}=\dfrac{5}{26}\)

c: \(=\dfrac{3}{4}\left(\dfrac{6}{11}+\dfrac{5}{11}\right)-\dfrac{1}{5}=\dfrac{3}{4}-\dfrac{1}{5}=\dfrac{11}{20}\)

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

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

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

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

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

\(-2\dfrac{1}{4}.\)\(\left(3\dfrac{5}{12}-1\dfrac{2}{9}\right)\)

=\(\dfrac{-9}{4}\).\(\left(\dfrac{41}{12}-\dfrac{11}{9}\right)\)

=\(\dfrac{-9}{4}.\dfrac{41}{12}-\dfrac{-9}{4}.\dfrac{11}{9}\)

=\(\dfrac{-123}{16}-\dfrac{-11}{4}\)

=\(\dfrac{-123}{16}-\dfrac{-44}{16}\)

=\(\dfrac{-79}{16}\)

\(\left(-25\%+0,75+\dfrac{7}{12}\right)\div\left(-2\dfrac{1}{8}\right)\)

=\(\left(\dfrac{-1}{4}+\dfrac{3}{4}+\dfrac{7}{12}\right)\div\left(\dfrac{-17}{8}\right)\)

=\(\left(\dfrac{-3}{12}+\dfrac{9}{12}+\dfrac{7}{12}\right).\dfrac{-8}{17}\)

=\(\dfrac{13}{12}.\dfrac{-8}{17}=\dfrac{-26}{51}\)

\(\dfrac{4}{7}.\dfrac{89}{5}-\dfrac{4}{5}.\dfrac{82}{7}\)

=\(\dfrac{4}{5}.\dfrac{89}{7}\dfrac{4}{5}.\dfrac{82}{7}\)

=\(\dfrac{4}{5}\left(\dfrac{89}{7}-\dfrac{82}{7}\right)\)

=\(\dfrac{4}{5}.1=\dfrac{4}{5}\)

\(\approx\) chúc bạn học tốt \(\approx\)

a. \(1\dfrac{4}{23}+\dfrac{5}{21}-\dfrac{4}{23}+0,5+\dfrac{16}{21}\)

\(=\left(1\dfrac{4}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{21}+\dfrac{16}{21}\right)+0,5\)

\(=1+1+0,5\)

\(=2,5\)

b. \(\dfrac{3}{7}.19\dfrac{1}{3}-\dfrac{3}{7}.33\dfrac{1}{3}\)

\(=\dfrac{3}{7}.\left(19\dfrac{1}{3}-33\dfrac{1}{3}\right)\)

\(=\dfrac{3}{7}.\left(-14\right)=-6\)

c. \(15\dfrac{1}{4}:\left(-\dfrac{5}{7}\right)-25\dfrac{1}{4}:\left(\dfrac{-5}{7}\right)\)

\(=\left(15\dfrac{1}{4}-25\dfrac{1}{4}\right):\left(-\dfrac{5}{7}\right)\)

\(=-10:\left(-\dfrac{5}{7}\right)\)

\(=14\)

d. \(\left(-\dfrac{2}{3}+\dfrac{3}{7}\right):\dfrac{4}{5}+\left(\dfrac{-1}{3}+\dfrac{4}{7}\right):\dfrac{4}{5}\)

\(=\dfrac{-5}{21}:\dfrac{4}{5}+\dfrac{5}{21}:\dfrac{4}{5}\)

\(=\left(\dfrac{-5}{7}+\dfrac{5}{7}\right):\dfrac{4}{5}\)

\(=0:\dfrac{4}{5}\)

\(=0\)

a,

\(1\dfrac{4}{23}+\dfrac{5}{21}-\dfrac{4}{23}+0,5+\dfrac{16}{21}\)

\(=\left(1\dfrac{4}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{21}+\dfrac{16}{21}\right)+0,5\)

\(=1+1-0,5=1,5\)

b,

\(\dfrac{3}{7}\cdot19\dfrac{1}{3}-\dfrac{3}{7}.33\dfrac{1}{3}\)

\(=\dfrac{3}{7}\left(19\dfrac{1}{3}-33\dfrac{1}{3}\right)=\dfrac{3}{7}.\left(-14\right)=-6\)

c,

\(15\dfrac{1}{4}:\left(-\dfrac{5}{7}\right)-25\dfrac{1}{4}:\left(-\dfrac{5}{7}\right)\)

\(=\left(15\dfrac{1}{4}-25\dfrac{1}{4}\right):\left(-\dfrac{5}{7}\right)=-10:\left(-\dfrac{5}{7}\right)=14\)

d,

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

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

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

\(=\left(-1+1\right):\dfrac{4}{5}=0:\dfrac{4}{5}=0\)

=-4/5(25+3/19-35-3/19)

=-4/5*(-10)

=8

\(25\dfrac{3}{19}:\left(-\dfrac{5}{4}\right)-35\dfrac{3}{19}:\left(-\dfrac{5}{4}\right)\\ =\left(25\dfrac{3}{19}-35\dfrac{3}{19}\right):\left(-\dfrac{5}{4}\right)\\ =-10:\left(-\dfrac{5}{4}\right)\\ =8\)