\(a.\)

\(-\dfrac{5}{9}\cdot\dfrac{12}{35}=\dfrac{\left(-5\right)\cdot12}{9\cdot35}=\dfrac{-60}{315}=-\dfrac{4}{21}\)

\(b.\)

\(\left(-\dfrac{5}{8}\right)\cdot-\dfrac{6}{55}=\dfrac{\left(-5\right)\cdot\left(-6\right)}{8\cdot55}=\dfrac{30}{440}=\dfrac{3}{44}\)

\(c.\)

\(\left(-7\right)\cdot\dfrac{2}{5}=-\dfrac{14}{5}\)

\(d.\)

\(-\dfrac{3}{8}\cdot\left(-6\right)=\dfrac{-3\cdot\left(-6\right)}{8}=\dfrac{18}{8}=\dfrac{9}{4}\)

a,-4/21 b,-3/44 c,-14/5 d,9/4

a) \(\dfrac{-5}{9}.\dfrac{12}{35}=\dfrac{-5.12}{9.35}=\dfrac{-4}{21}\) 

b) \(\dfrac{-5}{8}.\dfrac{-6}{55}=\dfrac{-5.-6}{8.55}=\dfrac{3}{44}\) 

c)\(-7.\dfrac{2}{5}=\dfrac{-7.2}{5}=\dfrac{-14}{5}\) 

d) \(\dfrac{-3}{8}.-6=\dfrac{-3.-6}{8}=\dfrac{9}{4}\)

a: Ta có: \(\dfrac{8}{9}-\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{72}\right)\)

\(=\dfrac{8}{9}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{3}-...+\dfrac{1}{8}-\dfrac{1}{9}\right)\)

=0

a)\(0,5+\dfrac{1}{3}+0,4+\dfrac{5}{7}+\dfrac{1}{6}-\dfrac{4}{35}\)

=\(\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}\right)+\left(\dfrac{2}{5}+\dfrac{5}{7}-\dfrac{4}{35}\right)=1+1=2\)

b) \(\dfrac{8}{9}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-\dfrac{1}{30}-\dfrac{1}{20}-\dfrac{1}{12}-\dfrac{1}{6}-\dfrac{1}{2}\)

=\(\dfrac{8}{9}-\left(\dfrac{1}{72}+\dfrac{1}{56}+\dfrac{1}{42}+\dfrac{1}{30}+\dfrac{1}{20}+\dfrac{1}{12}+\dfrac{1}{6}+\dfrac{1}{2}\right)\)

=\(\dfrac{8}{9}-\left(\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{2}\right)=\dfrac{8}{9}-\dfrac{8}{9}=0\)

a: \(=\left(\dfrac{3}{17}+\dfrac{14}{17}\right)+\left(\dfrac{-5}{13}-\dfrac{8}{13}\right)+\left(\dfrac{-18}{35}-\dfrac{17}{35}\right)\)

=1-1-1

=-1

b: \(=\dfrac{-3}{8}\left(\dfrac{1}{6}+\dfrac{5}{6}\right)+\dfrac{-5}{8}=\dfrac{-3}{8}-\dfrac{5}{8}=-1\)

c: \(=\dfrac{4}{4}\cdot\dfrac{5}{15}\cdot\dfrac{11}{11}=\dfrac{1}{3}\)

a) \(=\left(\dfrac{3}{17}+\dfrac{14}{17}\right)+\left(-\dfrac{5}{13}+-\dfrac{8}{13}\right)+\left(-\dfrac{18}{35}+-\dfrac{17}{35}\right)=1+-1+-1=-1\)

b) \(=-\dfrac{3}{8}\cdot\left(\dfrac{1}{6}+\dfrac{5}{6}\right)-\dfrac{10}{16}=-\dfrac{3}{8}-\dfrac{10}{16}=-1\)

c) \(=\left(-\dfrac{4}{11}\cdot-\dfrac{11}{4}\right)\cdot\dfrac{5}{15}=1\cdot\dfrac{1}{3}=\dfrac{1}{3}\)

a)\(=\left(-\dfrac{5}{13}+\dfrac{-8}{13}\right)+\left(-\dfrac{18}{35}-\dfrac{17}{35}\right)+\left(\dfrac{3}{14}+\dfrac{14}{17}\right)=-1-1+1=-1\)

b)\(=\dfrac{-3}{8}.\left(\dfrac{1}{6}+\dfrac{5}{6}\right)-\dfrac{10}{16}=-\dfrac{3}{8}.1-\dfrac{10}{16}=-\dfrac{6}{16}-\dfrac{10}{16}=-\dfrac{16}{16}=-1\)

c)\(\dfrac{-4.5.11}{11.5.3.-4}=\dfrac{1}{3}\)

Lời giải:

1.

$(\frac{5}{6})^{10}.(\frac{3}{10})^{10}=(\frac{5}{6}.\frac{3}{10})^{10}=(\frac{1}{4})^{10}$

$=\frac{1}{4^{10}}$

2.

$(\frac{4}{7})^{19}: (\frac{-12}{35})^{19}=(\frac{4}{7}: \frac{-12}{35})^{19}=(\frac{-5}{3})^{19}$

3.

$(\frac{-3}{7})^7:\frac{-3}{5}=\frac{(-3)^7}{7^7}.\frac{5}{-3}=\frac{5.(-3)^6}{7^7}=\frac{5.3^6}{7^7}$

giúp mình với ạ

1) \(\left(\dfrac{5}{6}\right)^{10}\cdot\left(\dfrac{3}{10}\right)^{10}\)

\(=\left(\dfrac{5}{6}\cdot\dfrac{3}{10}\right)^{10}\)

\(=\left(\dfrac{1}{4}\right)^{10}\)

2) \(\left(\dfrac{4}{9}\right)^{19}:\left(\dfrac{-12}{35}\right)^{19}\)

\(=\left(\dfrac{4}{9}:\dfrac{-12}{35}\right)^{19}\)

\(=\left(\dfrac{4}{9}\cdot\dfrac{35}{-12}\right)^{19}\)

\(=\left(-\dfrac{35}{27}\right)^{19}\)

3) \(\left(\dfrac{-3}{7}\right)^7:\left(\dfrac{-3}{5}\right)^7\)

\(=\left(\dfrac{-3}{7}:\dfrac{-3}{5}\right)^7\)

\(=\left(\dfrac{-3}{7}\cdot\dfrac{5}{-3}\right)^7\)

\(=\left(\dfrac{5}{7}\right)^7\)

a >

b <

c >

d <

>

<

>

<

\(a:ta.c\text{ó}:BCNN:12\\ \dfrac{2}{3}=\dfrac{2\cdot4}{3\cdot4}=\dfrac{8}{12};\dfrac{1}{4}=\dfrac{1\cdot3}{4\cdot3}=\dfrac{3}{12}\\ v\text{ì }\dfrac{8}{12}< \dfrac{3}{12}n\text{ê}n\dfrac{2}{3}< \dfrac{1}{4}\\ b:ta.c\text{ó}:\\ 10=2\cdot5\\ 8=2^3\\ \Rightarrow BCNN=2^3\cdot5=8\cdot5=40\\ \dfrac{7}{10}=\dfrac{7\cdot4}{10\cdot4}=\dfrac{28}{40};\dfrac{7}{8}=\dfrac{7\cdot5}{8\cdot5}=\dfrac{35}{40}\\ v\text{ì }\dfrac{28}{40}< \dfrac{35}{40}n\text{ê}n\dfrac{7}{10}< \dfrac{7}{8}\\ c:ta.c\text{ó}:\\ 7=7;5=5\\ \Rightarrow BCNN=7\cdot5=35\\ \dfrac{6}{7}=\dfrac{6\cdot5}{7\cdot5}=\dfrac{30}{35};\dfrac{3}{5}=\dfrac{3\cdot7}{5\cdot7}=\dfrac{21}{35}\\ v\text{ì }\dfrac{30}{35}>\dfrac{21}{35}n\text{ê}n\dfrac{6}{7}>\dfrac{3}{5}\\ d:ta.c\text{ó}:\\ 21=3\cdot7\\ 72=2^3\cdot3^2\\ \Rightarrow BCNN=2^3\cdot3^2\cdot7=504\\ \dfrac{14}{21}=\dfrac{14\cdot24}{21\cdot24}=\dfrac{336}{504};\dfrac{60}{72}=\dfrac{60\cdot7}{72\cdot7}=\dfrac{420}{504}\\ v\text{ì }\dfrac{336}{504}< \dfrac{420}{504}n\text{ê}n\dfrac{14}{21}< \dfrac{60}{72}\)

1: \(\dfrac{1}{2}+\dfrac{9}{10}+\dfrac{5}{6}-\dfrac{11}{14}-\dfrac{1}{3}+\dfrac{-4}{35}\)

\(=\left(\dfrac{1}{2}+\dfrac{5}{6}-\dfrac{1}{3}\right)+\dfrac{9}{10}-\left(\dfrac{11}{14}+\dfrac{4}{35}\right)\)

\(=\dfrac{3+5-2}{6}+\dfrac{9}{10}-\dfrac{55+8}{70}\)

\(=1+\dfrac{9}{10}-\dfrac{9}{10}\)

=1

a: \(\dfrac{-1}{2}+\dfrac{5}{6}+\dfrac{1}{3}\)

\(=\dfrac{-3}{6}+\dfrac{5}{6}+\dfrac{2}{6}\)

\(=\dfrac{4}{6}=\dfrac{2}{3}\)

b: \(\dfrac{-3}{8}+\dfrac{7}{4}-\dfrac{1}{12}\)

\(=\dfrac{-9}{24}+\dfrac{42}{24}-\dfrac{2}{24}\)

\(=\dfrac{31}{24}\)

c: \(\dfrac{3}{5}:\left(\dfrac{1}{4}\cdot\dfrac{7}{5}\right)=\dfrac{3}{4}:\dfrac{7}{20}=\dfrac{3}{4}\cdot\dfrac{20}{7}=\dfrac{15}{7}\)

d: \(\dfrac{10}{11}+\dfrac{4}{11}:4-\dfrac{1}{8}\)

\(=\dfrac{10}{11}+\dfrac{1}{11}-\dfrac{1}{8}=\dfrac{7}{8}\)

6/21-(−12/44)+10/14−(1/(−4))−18/33

=2/7+12/44+5/7−((−1)/4)−6/11=2/7+12/44+5/7−((−1)/4)−6/11

=2/7+3/11+5/7+1/4−6/11=2/7+3/11+5/7+1/4−6/11

=(3/11−6/11)+(2/7+5/7)+1/4=(3/11−6/11)+(2/7+5/7)+1/4

=−3/11+7/7+1/4=−3/11+7/7+1/4

=43/44

a: \(=\dfrac{3}{8}\left(27+\dfrac{1}{5}-51-\dfrac{1}{5}\right)+19\)

\(=-24\cdot\dfrac{3}{8}+19=-9+19=10\)

b: \(=\left(35+\dfrac{1}{6}-46-\dfrac{1}{6}\right):\left(\dfrac{-4}{5}\right)\)

\(=\dfrac{-11\cdot5}{-4}=\dfrac{55}{4}\)

c: \(=\left(\dfrac{-15+8}{20}\right):\left[\dfrac{3}{7}+\dfrac{7}{3}\cdot\dfrac{12-5}{20}\right]\)

\(=\dfrac{-7}{20}:\left(\dfrac{3}{7}+\dfrac{49}{60}\right)\)

\(=-\dfrac{147}{523}\)