a) BCNN(3,11)=33

Thừa số phụ: 33 : 3 = 11; 33 : 11 = 3

\(\begin{array}{l}\dfrac{5}{3} = \dfrac{{5.11}}{{3.11}} = \dfrac{{55}}{{33}}\\\dfrac{{15}}{{11}} = \dfrac{{15.3}}{{11.3}} = \dfrac{{45}}{{33}}\end{array}\)

Vì \(\dfrac{{55}}{{33}} > \dfrac{{45}}{{33}}\)

Nên \(\dfrac{5}{3}kg > \dfrac{{15}}{{11}}\)kg

b) BCNN(6,5)=30

Thừa số phụ: 30 : 6 = 5; 30 : 5 = 6

\(\begin{array}{l}\dfrac{5}{6} = \dfrac{{5.5}}{{6.5}} = \dfrac{{25}}{{30}}\\\dfrac{4}{5} = \dfrac{{4.6}}{{5.6}} = \dfrac{{24}}{{30}}\end{array}\)

Vì \(\dfrac{{25}}{{30}} > \dfrac{{24}}{{30}}\)

Nên \(\dfrac{5}{6}km/h > \dfrac{4}{5}km/h\) 

a)\(\dfrac{2}{3}giờ< \dfrac{3}{4}giờ\)

b)\(\dfrac{7}{10}m< \dfrac{3}{4}m\)

c)\(\dfrac{7}{8}kg< \dfrac{9}{10}kg\)

d)\(\dfrac{5}{6}\)km/h >\(\dfrac{7}{9}\)km/h

a) h < h

b) m < m

c) kg < kg

d) km/h > km/h.

a)

\(\dfrac{5}{8}\left(kg\right)< \dfrac{6}{8}\left(kg\right)\)

Do 13 > 12 nên \(\dfrac{13}{12}>1\)

hay: \(\dfrac{13}{12}\left(kg\right)>1\left(kg\right)\)

b)

\(\dfrac{11}{12}\left(l\right)>\dfrac{11}{14}\left(l\right)\)

Có:

\(\dfrac{5}{3}=\dfrac{5\times3}{3\times3}=\dfrac{15}{9}\)

nên: \(\dfrac{5}{3}\left(l\right)=\dfrac{15}{9}\left(l\right)\)

c)

\(\dfrac{5}{6}=\dfrac{5\times3}{6\times3}=\dfrac{15}{18n}\)

nên \(\dfrac{5}{6}\left(m\right)< \dfrac{17}{8}\left(m\right)\)

Có: \(2=\dfrac{2\times7}{7}=\dfrac{14}{7}\)

nên: \(\dfrac{16}{7}\left(m\right)>2\left(m\right)\)

a, <

b,>

c,<

d,>

e,=

g,>

`3/5` giờ `=0,6` giờ

`5/4m=1,25m`

`5/8` giờ `=0,625` giờ

`9/5km=1,8km`

`3/5` phút `=0,6` giờ

`7/8kg=0,875kg`

\(\dfrac{3}{5}\) giờ = 0,6 giờ
\(\dfrac{5}{4}m=1,25m\)
\(\dfrac{5}{8}\) giờ = 0,625 giờ
\(\dfrac{9}{5}km=1,8km\)
\(\dfrac{3}{5}\) phút = 0,6 phút
\(\dfrac{7}{8}kg=0,875kg\)

a) \(=\left(13\dfrac{2}{7}+2\dfrac{5}{7}\right):\left(-\dfrac{8}{9}\right)\)

\(=16:\dfrac{-8}{9}=\dfrac{-8\cdot\left(-2\right)\cdot9}{-8}=-18\)

b) 

\(=\left(\dfrac{-6}{11}\cdot\dfrac{11}{-6}\right)\cdot\dfrac{7\cdot10\cdot\left(-2\right)}{10}\)

\(=-14\)

c) \(=\dfrac{-1}{2}\cdot\dfrac{4}{3}\cdot\dfrac{-7}{2}\)

\(=\dfrac{-1\cdot2\cdot2\cdot\left(-7\right)}{2\cdot3\cdot2}=\dfrac{7}{3}\)

c) \(\dfrac{11}{10}-\dfrac{-7}{24}=\dfrac{11}{10}+\dfrac{7}{24}=\dfrac{167}{120}\)

e) \(\dfrac{-8}{3}\cdot\dfrac{15}{7}=\dfrac{-120}{21}=\dfrac{-40}{7}\)

f) \(\dfrac{-2}{5}\cdot4\dfrac{1}{2}=\dfrac{-2}{5}\cdot\dfrac{9}{2}=-\dfrac{9}{5}\)

g) \(\dfrac{5}{3}:\dfrac{5}{-3}=\dfrac{5}{3}:\dfrac{-5}{3}=\dfrac{5}{3}\cdot\dfrac{-3}{5}=-1\)

h) \(\dfrac{5}{4}:\left(-9\right)=\dfrac{5}{4}:\dfrac{-9}{1}=\dfrac{5}{4}\cdot\dfrac{-1}{9}=-\dfrac{5}{36}\)

*Trả lời :

a) \(-\dfrac{3}{4}.5\dfrac{3}{13}-0,75.\dfrac{36}{13}\)

= \(-\dfrac{3}{4}.\dfrac{68}{13}-\dfrac{3}{4}.\dfrac{36}{13}\)

=\(\dfrac{3}{4}.\dfrac{-68}{13}-\dfrac{3}{4}.\dfrac{36}{13}\)

=\(\dfrac{3}{4}.\cdot\left(\dfrac{-68}{13}-\dfrac{36}{13}\right)\)

=\(\dfrac{3}{4}.\left(-8\right)\)

= \(-6\)

b)\(4\dfrac{5}{9}:\left(-\dfrac{5}{7}\right)+\dfrac{49}{9}:\left(-\dfrac{5}{7}\right)\)

=\(\dfrac{41}{9}-\left(-\dfrac{5}{7}\right)+\dfrac{49}{9}:\left(-\dfrac{5}{7}\right)\)

=\(\left(\dfrac{41}{9}+\dfrac{49}{9}\right):\left(-\dfrac{5}{7}\right)\)

=\(\dfrac{90}{9}:\left(-\dfrac{5}{7}\right)\)

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

=\(-14\)

c)\(\left(-\dfrac{3}{5}+\dfrac{4}{9}\right):\dfrac{7}{11}+\left(-\dfrac{2}{5}+\dfrac{5}{9}\right):\dfrac{7}{11}\)

=\(\left(-\dfrac{3}{5}\right)+\dfrac{4}{9}:\dfrac{7}{11}+\left(-\dfrac{2}{5}\right)+\dfrac{5}{9}:\dfrac{7}{11}\)(áp dụng tính chất phá ngoặc )

=\(\left\{\left[-\dfrac{3}{5}+\left(-\dfrac{2}{5}\right)\right]+\left(\dfrac{4}{9}+\dfrac{5}{9}\right)\right\}:\dfrac{7}{11}\)

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

=\(\left(-1+1\right):\dfrac{7}{11}\)

\(=0:\dfrac{7}{11}\)

=0.

d)\(\dfrac{6}{7}:\left(\dfrac{3}{26}-\dfrac{3}{13}\right)+\dfrac{6}{7}:\left(\dfrac{1}{10}-\dfrac{8}{5}\right)\)

=\(\dfrac{6}{7}:\left[\dfrac{3}{26}+\left(-\dfrac{6}{26}\right)\right]+\dfrac{6}{7}:\left[\dfrac{1}{10}+\left(-\dfrac{16}{10}\right)\right]\)

=\(\dfrac{6}{7}:\left(-\dfrac{3}{26}\right)+\dfrac{6}{7}:\left(-\dfrac{3}{2}\right)\)

=\(\dfrac{6}{7}:\left[\left(-\dfrac{3}{26}\right)+\left(-\dfrac{39}{26}\right)\right]\)

=\(\dfrac{6}{7}:\left(-\dfrac{21}{13}\right)\)

=\(-\dfrac{26}{49}\)

giúp mik với mọi người

a: \(=\dfrac{4\cdot2+4\cdot9}{55}+\dfrac{5}{6}=\dfrac{4}{5}+\dfrac{5}{6}=\dfrac{49}{30}\)

b: \(=\dfrac{3}{2}\cdot\dfrac{3}{5}-\left(\dfrac{3}{7}+\dfrac{3}{20}\right)\cdot\dfrac{10}{3}\)

\(=\dfrac{9}{10}-\dfrac{81}{140}\cdot\dfrac{10}{3}\)

\(=\dfrac{9}{10}-\dfrac{27}{14}=\dfrac{-36}{35}\)

c: \(=15+\dfrac{3}{13}-3-\dfrac{4}{7}-8-\dfrac{3}{13}\)

\(=4-\dfrac{4}{7}=\dfrac{24}{7}\)

d: \(=\dfrac{-7}{9}\left(\dfrac{4}{11}+\dfrac{7}{11}\right)+5+\dfrac{7}{9}=5\)

\(a.\left[-\dfrac{6}{11}.\dfrac{11}{-6}\right].\dfrac{7}{10}.\left(-20\right)=1.7.\left(-2\right)=-14\)

\(b.\dfrac{-1}{2}:\dfrac{3}{4}.\dfrac{-7}{2}=\dfrac{7}{4}:\dfrac{3}{4}=\dfrac{7}{3}\)

\(c.\dfrac{93}{7}:-\dfrac{8}{9}+\dfrac{19}{7}:\dfrac{-8}{9}=\left(\dfrac{93}{7}+\dfrac{19}{7}\right):-\dfrac{8}{9}=\dfrac{-9}{8}.\dfrac{112}{7}=-18\)

e: \(D=\dfrac{-10}{12}-\dfrac{7}{12}-\dfrac{4}{12}=\dfrac{-21}{12}=-\dfrac{7}{4}\)

f: \(F=\dfrac{-27}{36}+\dfrac{12}{36}+\dfrac{10}{36}=\dfrac{-5}{36}\)

g: \(G=\dfrac{209}{99}+\dfrac{36}{99}+\dfrac{66}{99}=\dfrac{311}{99}\)

h: \(H=\dfrac{10}{24}-\dfrac{42}{24}+\dfrac{3}{24}=-\dfrac{29}{24}\)

\(D=\dfrac{-5}{6}+\dfrac{-7}{12}-\dfrac{1}{3}=-\dfrac{7}{4}\)

\(F=\dfrac{-3}{4}+\dfrac{1}{3}-\dfrac{-5}{18}=-\dfrac{5}{36}\)

\(G=\dfrac{19}{9}-\dfrac{-4}{11}+\dfrac{2}{3}=\dfrac{311}{99}\)

\(H=\dfrac{5}{12}+\dfrac{-7}{4}-\dfrac{1}{-8}=-\dfrac{29}{24}\)

\(e,D=\dfrac{-5}{6}+\dfrac{-7}{12}-\dfrac{1}{3}=\dfrac{-10}{12}+\dfrac{-7}{12}-\dfrac{4}{12}=\dfrac{\left(-10\right)+\left(-7\right)-4}{12}=\dfrac{-21}{12}=\dfrac{-7}{4}\\ f,F=\dfrac{-3}{4}+\dfrac{1}{3}-\dfrac{-5}{18} =\dfrac{-27}{36}+\dfrac{12}{36}-\dfrac{-10}{36}=\dfrac{\left(-27\right)+12-\left(-10\right)}{36}=\dfrac{-5}{36}\)

\(g,G=\dfrac{19}{9}-\dfrac{-4}{11}+\dfrac{2}{3}=\dfrac{209}{99}-\dfrac{-36}{99}+\dfrac{66}{99}=\dfrac{209-\left(-36\right)+66}{99}=\dfrac{311}{99}\\ h,H=\dfrac{5}{12}+\dfrac{-7}{4}-\dfrac{1}{-8}=\dfrac{5}{12}-\dfrac{7}{4}+\dfrac{1}{8}=\dfrac{10}{24}-\dfrac{42}{24}+\dfrac{3}{24}=\dfrac{10-42+3}{24}=\dfrac{-29}{24}\)