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a, 3 \(\frac{14}{19}\)+ \(\frac{13}{17}\)+ \(\frac{35}{43}\)+ 6\(\frac{5}{19}\)+ \(\frac{8}{43}\)= \(\left(3\frac{14}{19}+6\frac{5}{19}\right)+\left(\frac{35}{43}+\frac{8}{43}\right)+\frac{13}{17}=\)\(9+1+\frac{13}{17}=8+\frac{13}{17}=8\frac{13}{17}\)
b, \(\frac{-5}{7}.\frac{2}{11}+\frac{-5}{7}.\frac{9}{11}+1\frac{5}{7}\)\(=\frac{-5}{7}\left(\frac{2}{11}+\frac{9}{11}\right)+1\frac{5}{7}\)\(=\frac{-5}{7}.1+1\frac{5}{7}\)\(=\frac{-5}{7}+\frac{12}{7}=\frac{7}{7}=1\)
Chúc bn học tốt
\(3\frac{14}{19}+\frac{13}{17}+\frac{35}{43}+6\frac{5}{19}+\frac{8}{43}\)
\(=\left(3\frac{14}{19}+6\frac{5}{19}\right)+\left(\frac{35}{43}+\frac{8}{43}\right)+\frac{13}{17}\)
\(=10+1+\frac{13}{17}=11+\frac{13}{17}=11\frac{13}{17}\)

số thứ nhất là
20 :(2+3) x 2 =8 (đơn vị)
số thứ hai là
20 - 8 = 12 (đơn vị)
vậy 2 số đó số đó là 8 và 12
chọn A. 8 và 12
câu 2 a)\(\frac{-11}{12}\) và \(\frac{-17}{18}\)
mẫu số chung của 12 và 18 là 36
=> \(-\frac{33}{36}\) và\(\frac{-34}{36}\)
vì \(\frac{33}{36}<\frac{34}{36}\) =>\(-\frac{33}{36}>-\frac{34}{36}\)
=>\(\frac{-11}{12}\) <\(\frac{-17}{18}\)

\(3\frac{14}{19}+\frac{13}{17}+\frac{35}{43}+6\frac{5}{19}+\frac{8}{43}\)
\(=\left(3\frac{14}{19}+6\frac{5}{19}\right)+\left(\frac{35}{43}+\frac{8}{43}\right)+\frac{13}{17}\)
\(=10+1+\frac{13}{17}=11+\frac{13}{17}=11\frac{13}{17}\)
\(\frac{-5}{7}.\frac{2}{11}+\frac{-5}{7}.\frac{9}{11}+1\frac{5}{7}\)
\(=\frac{-5}{7}.\frac{2}{11}+\frac{-5}{7}.\frac{9}{11}+1+\frac{-5}{7}.\left(-1\right)\)
\(=\frac{-5}{7}\left(\frac{2}{11}+\frac{9}{11}-1\right)+1\)
\(=\frac{-5}{7}.0+1==0+1=1\)

Bài 1:
a) Ta có: \(\frac{3}{5}+\frac{4}{15}\)
\(=\frac{9}{15}+\frac{4}{15}\)
\(=\frac{13}{15}\)
b) Ta có: \(\frac{-3}{5}+\frac{5}{7}\)
\(=\frac{-21}{35}+\frac{25}{35}=\frac{4}{35}\)
c) Ta có: \(\frac{5}{6}:\frac{-7}{12}\)
\(=\frac{5}{6}\cdot\frac{-12}{7}=\frac{-60}{42}=\frac{-10}{7}\)
d) Ta có: \(\frac{-21}{24}:\frac{-14}{8}\)
\(=\frac{-7}{8}:\frac{-7}{4}\)
\(=\frac{-7}{8}\cdot\frac{4}{-7}=\frac{4}{8}=\frac{1}{2}\)
e) Ta có: \(\frac{-3}{5}\cdot\frac{5}{7}+\frac{-3}{5}\cdot\frac{3}{7}+\frac{-3}{5}\cdot\frac{6}{7}\)
\(=\frac{-3}{5}\left(\frac{5}{7}+\frac{3}{7}+\frac{6}{7}\right)\)
\(=-\frac{3}{5}\cdot2=\frac{-6}{5}\)
f) Ta có: \(\frac{1}{3}\cdot\frac{4}{5}+\frac{1}{3}\cdot\frac{6}{5}-\frac{4}{3}\)
\(=\frac{1}{3}\cdot\frac{4}{5}+\frac{1}{3}\cdot\frac{6}{5}-\frac{1}{3}\cdot4\)
\(=\frac{1}{3}\left(\frac{4}{5}+\frac{6}{5}-4\right)\)
\(=\frac{1}{3}\cdot\left(-2\right)=\frac{-2}{3}\)
g) Ta có: \(\frac{4}{19}\cdot\frac{-3}{7}+\frac{-3}{7}\cdot\frac{5}{19}+\frac{5}{7}\)
\(=\frac{4}{19}\cdot\frac{-3}{7}+\frac{5}{19}\cdot\frac{-3}{7}+\frac{-3}{7}\cdot\frac{5}{-3}\)
\(=-\frac{3}{7}\left(\frac{4}{19}+\frac{5}{19}+\frac{-5}{3}\right)\)
\(=\frac{-3}{7}\cdot\left(\frac{27}{57}+\frac{-95}{57}\right)\)
\(=\frac{-3}{7}\cdot\frac{-68}{57}=\frac{68}{133}\)
h) Ta có: \(\frac{5}{9}\cdot\frac{7}{13}+\frac{5}{9}\cdot\frac{9}{13}-\frac{5}{9}\cdot\frac{3}{13}\)
\(=\frac{5}{9}\left(\frac{7}{13}+\frac{9}{13}-\frac{5}{13}\right)\)
\(=\frac{5}{9}\)

\(\frac{-7}{31}\) và \(\frac{6}{31}\)
\(\frac{-7}{31}<0;\frac{6}{31}>0\)
=>\(-\frac{7}{31}<\frac{6}{31}\)
\(\frac{-97}{128}\) và \(-\frac{99}{128}\)
vì \(\frac{97}{128}<\frac{99}{128}\) =>\(\frac{-97}{128}>-\frac{99}{128}\)
\(\frac37\) và \(\frac{-6}{7}\)
vì\(\frac37>0;-\frac67<0\)
=>\(\frac37>-\frac67\)

\(\frac37+\frac{5}{13}+\frac{4}{13}\)
=\(\frac37+\left(\frac{5}{13}+\frac{4}{13}\right)\)
=\(\frac37+\frac{9}{13}\)
=\(\frac{39}{91}+\frac{63}{91}\)
=\(\frac{102}{91}\)
\(\left(\frac38+\frac{-3}{4}+\frac{7}{12}\right):\frac56+\frac12\)
=\(\left(\frac{9}{24}+\frac{-18}{24}+\frac{7}{24}\right)\times\frac65+\frac12\)
=\(\frac{5}{24}\times\frac65+\frac12\)
=\(\frac14+\frac12\)
=\(\frac14+\frac24\)
=\(\frac34\)

Bài 1:
a) Ta có: \(\frac{8}{40}+\frac{-4}{20}-\frac{3}{5}\)
\(=\frac{1}{5}+\frac{-1}{5}-\frac{3}{5}\)
\(=\frac{-3}{5}\)
b) Ta có: \(\frac{-7}{12}+\frac{-2}{12}-\frac{-3}{36}\)
\(=\frac{-7}{12}+\frac{-2}{12}-\frac{-1}{12}\)
\(=\frac{-9+1}{12}=\frac{-8}{12}=\frac{-2}{3}\)
c) Ta có: \(\left(\frac{1}{6}+\frac{-4}{13}\right)-\left(-\frac{17}{6}-\frac{30}{13}\right)\)
\(=\frac{1}{6}+\frac{-4}{13}+\frac{17}{6}+\frac{30}{13}\)
\(=3+2=5\)
d) Ta có: \(-\frac{-5}{4}+\frac{7}{4}-\frac{-11}{7}+\frac{2}{7}\)
\(=\frac{5}{4}+\frac{7}{4}+\frac{11}{7}+\frac{2}{7}\)
\(=3+\frac{13}{7}=\frac{21}{7}+\frac{13}{7}=\frac{34}{7}\)
e) Ta có: \(-\frac{1}{8}+\frac{-7}{9}+\frac{-7}{8}+\frac{6}{7}+\frac{2}{14}\)
\(=-1+1+\frac{-7}{9}\)
\(=-\frac{7}{9}\)
f) Ta có: \(\frac{-2}{9}-\frac{11}{-9}+\frac{5}{7}-\frac{-6}{-7}\)
\(=\frac{-2-\left(-11\right)}{9}+\frac{5-6}{7}\)
\(=1+\frac{-1}{7}=\frac{7}{7}+\frac{-1}{7}=\frac{6}{7}\)

a)\(\left(\frac38+\frac{-3}{4}+\frac{7}{12}\right):\frac56+\frac12\)
=\(\left(\frac{9}{24}+\frac{-18}{24}+\frac{14}{24}\right)\times\frac65+\frac12\)
=\(\frac{5}{24}\times\frac65+\frac12\)
=\(\frac14+\frac12\)
=\(\frac14+\frac24\)
=\(\frac34\)
b)\(\frac12+\frac34-\left(\frac34-\frac45\right)\)
=\(\frac12+\frac34-\frac34+\frac45\)
=\(\left(\frac12+\frac45\right)+\left(\frac34-\frac34\right)\)
=\(\left(\frac{5}{10}+\frac{8}{10}\right)+0\)
=\(\frac{13}{10}\)
Bài làm
a) \(-\frac{3}{7}+\frac{3}{4}:\frac{3}{14}\)
= \(-\frac{3}{7}+\frac{3}{4}.\frac{14}{3}\)
= \(-\frac{3}{7}+\frac{7}{2}\)
\(=-\frac{7}{14}+\frac{49}{14}\)
\(=\frac{42}{14}=3\)
b) \(5-\frac{7}{39}:\frac{7}{13}+\frac{8}{9}:4\)
\(=5=\frac{7}{39}.\frac{13}{7}+\frac{8}{9}.\frac{1}{4}\)
\(=5-\frac{1}{3}+\frac{2}{9}\)
\(=\frac{45}{9}-\frac{3}{9}+\frac{2}{9}\)
\(=\frac{44}{9}\)
c) \(\left(\frac{5}{12}:\frac{11}{6}+\frac{5}{12}:\frac{11}{5}\right)-\frac{-7}{12}\)
\(=\left(\frac{5}{12}.\frac{6}{11}+\frac{5}{12}.\frac{5}{11}\right)+\frac{7}{12}\)
\(=\left[\frac{5}{12}\left(\frac{6}{11}+\frac{5}{11}\right)\right]+\frac{7}{12}\)
\(=\frac{5}{12}+\frac{7}{12}\)
\(=\frac{12}{12}=1\)
d) \(-\frac{5}{9}+\frac{14}{9}\left(\frac{3}{4}-\frac{2}{5}\right):49\)
\(=-\frac{5}{9}+\frac{14}{9}\left(\frac{15}{20}-\frac{8}{20}\right):49\)
\(=-\frac{5}{9}+\frac{14}{9}.\frac{7}{20}.\frac{1}{49}\)
\(=-\frac{5}{9}+\frac{7}{9}.\frac{7}{10}.\frac{1}{7.7}\)
\(=-\frac{5}{9}+\frac{1}{90}\)
\(=-\frac{50}{90}+\frac{1}{90}=-\frac{49}{90}\)