\(a,A=\dfrac{\dfrac{5}{4}+\dfrac{5}{5}+\dfrac{5}{7}-\dfrac{5}{11}}{\dfrac{10}{4}+\dfrac{10}{5}+\dfrac{10}{7}-\dfrac{10}{11}}\\ =\dfrac{5.\left(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}{10.\left(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}\\ =\dfrac{5}{10}\\ =\dfrac{1}{2}\)

Vậy \(A=\dfrac{1}{2}\)

\(b,B=\dfrac{2+\dfrac{6}{5}-\dfrac{6}{7}-\dfrac{6}{11}}{\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}}\\ =\dfrac{3.\left(\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}\right)}{\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}}\\ =3\)

Vậy \(B=3\)

`a)1/7xx2/7+1/7xx5/7+6/7`

`=1/7xx(2/7+5/7)+6/7`

`=1/7xx1+6/7`

`=1/7+6/7=1`

`b)6/11xx4/9+6/11xx7/9-6/11xx2/9`

`=6/11xx(4/9+7/9-2/9)`

`=6/11xx9/9`

`=6/11`

Sorry nãy ghi thiếu.

`c)4/25xx5/8xx25/4xx24`

`=(4xx5xx25xx24)/(25xx8xx4)`

`=(4xx5xx24)/(4xx8)`

`=(5xx24)/8`

`=5xx3=15`

a, \(\dfrac{1}{7}.\dfrac{2}{7}+\dfrac{1}{7}.\dfrac{5}{7}+\dfrac{6}{7}\)

\(=\dfrac{1}{7}.\left(\dfrac{2}{7}+\dfrac{5}{7}\right)+\dfrac{6}{7}\)
\(=\dfrac{1}{7}.1+\dfrac{6}{7}\)

\(=\dfrac{1}{7}+\dfrac{6}{7}=1\)

b, \(\dfrac{6}{11}.\dfrac{4}{9}+\dfrac{6}{11}.\dfrac{7}{9}-\dfrac{6}{11}.\dfrac{2}{9}\)

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

\(=\dfrac{6}{11}.1=\dfrac{6}{11}\)

c, \(\dfrac{4}{25}.\dfrac{5}{8}.\dfrac{25}{4}.24\)

\(=\left(\dfrac{4}{25}.\dfrac{25}{4}\right).\left(\dfrac{5}{8}.24\right)\)

\(=1.15=15\)

a: \(=\dfrac{-6}{11}:\dfrac{3\cdot11}{4\cdot5}=\dfrac{-6}{11}\cdot\dfrac{20}{33}=\dfrac{-2}{11}\cdot\dfrac{20}{11}=\dfrac{-40}{121}\)

b: \(=\dfrac{7}{12}+\dfrac{5}{72}-\dfrac{11}{36}=\dfrac{42}{72}+\dfrac{5}{72}-\dfrac{22}{72}=\dfrac{25}{72}\)

c: \(=\dfrac{13}{10}:\dfrac{-5}{13}=\dfrac{-169}{50}\)

L: \(\dfrac{7}{6}\) +\(\dfrac{8}{6}\) = \(\dfrac{15}{6}\)

T: \(\dfrac{1}{11}\) + \(\dfrac{7}{11}\) + \(\dfrac{9}{11}\) = \(\dfrac{17}{11}\)

Ạ: \(\dfrac{4}{11}\) + \(\dfrac{2}{11}\) = \(\dfrac{6}{11}\)

Đ: \(\dfrac{3}{6}\) + \(\dfrac{1}{6}\) + \(\dfrac{7}{6}\) = \(\dfrac{11}{6}\)

À: \(\dfrac{3}{11}\) + \(\dfrac{9}{11}\) = \(\dfrac{12}{11}\)

ĐÀ LẠT

a) 3/4 + (-7/5) + 1/4 + (-3/5)

= (3/4 + 1/4) + (-7/5 - 3/5)

= 1 - 2

= -1

b) 4/9 . 7/11 - 4/11 . 2/9 + 6/11 . 4/9

= 4/9 . (7/11 - 2/11 + 6/11)

= 4/9 . 1

= 4/9

a) 

\(\dfrac{3}{4}+\dfrac{-7}{5}+\dfrac{1}{4}+\dfrac{-3}{5}\)

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

\(=1-2\)

\(=-1\)

b)

\(\dfrac{4}{9}.\dfrac{7}{11}-\dfrac{4}{11}.\dfrac{2}{9}+\dfrac{6}{11}.\dfrac{4}{9}\)

\(=\dfrac{28}{99}-\dfrac{8}{99}+\dfrac{24}{99}\)

\(=\dfrac{28-8+24}{99}\)

\(=\dfrac{44}{99}\)

\(=\dfrac{4}{9}\)

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,>

a)

b)

+) Quy đồng mẫu số ba phân số $\frac{1}{4};\frac{3}{4};\frac{5}{8}$

$\frac{1}{4} = \frac{{1 \times 2}}{{4 \times 2}} = \frac{2}{8}$

$\frac{3}{4} = \frac{{3 \times 2}}{{4 \times 2}} = \frac{6}{8}$ ; Giữ nguyên phân số $\frac{5}{8}$

Vì $\frac{2}{8} < \frac{5}{8} < \frac{6}{8}$ nên $\frac{1}{4} < \frac{5}{8} < \frac{3}{4}$

Vậy các phân số xếp theo thứ tự từ bé đến lớn là: $\frac{1}{4};\,\,\frac{5}{8};\,\,\frac{3}{4}$

+) Quy đồng mẫu số ba phân số $\frac{2}{3};\,\,\frac{2}{9};\,\,\frac{5}{9}$

$\frac{2}{3} = \frac{{2 \times 3}}{{3 \times 3}} = \frac{6}{9}$ ; Giữ nguyên phân số $\frac{2}{9}$; $\frac{5}{9}$

Vì $\frac{2}{9} < \frac{5}{9} < \frac{6}{9}$ nên $\frac{2}{9} < \frac{5}{9} < \frac{2}{3}$

Vậy các phân số xếp theo thứ tự từ bé đến lớn là $\frac{2}{9};\,\,\frac{5}{9};\,\,\frac{2}{3}$

sai môn r bn