a: \(=\dfrac{-28}{36}+\dfrac{15}{36}-\dfrac{26}{36}=\dfrac{-39}{36}=\dfrac{-13}{12}\)

b: \(=\dfrac{11}{9}\left(\dfrac{15}{4}-\dfrac{7}{4}-\dfrac{5}{4}\right)=\dfrac{11}{9}\cdot\dfrac{3}{4}=\dfrac{11}{12}\)

c: \(=15+\dfrac{9}{7}+6+\dfrac{2}{3}-5-\dfrac{5}{9}\)

\(=16+\dfrac{88}{63}=\dfrac{1096}{63}\)

d: \(=\dfrac{5}{6}-\dfrac{1}{3}+\dfrac{2}{18}\)

\(=\dfrac{15-6+2}{18}=\dfrac{11}{18}\)

sao bn phũ với mk thế:(( đx ko giải lại còn nói thế

Nó dễ mà :(

câu a
\(A=\frac{33.10^3}{2^3.5.10^3+7000}=\frac{33.10^3}{2^3.5.10^3+7.10^3}=\frac{33.10^3}{10^3\left(2^3.5+7\right)}=\frac{33.10^3}{10^3.47}=\frac{33}{47}\)
\(B=\frac{3774}{5217}=\frac{34.111}{47.111}=\frac{34}{47}\)
\(\Rightarrow\frac{33}{47}< \frac{34}{47}\)
=> A<B

Đề răng dài thế này thì tui giải từng câu hí

a) \(\dfrac{-7}{9}+\dfrac{5}{12}-\dfrac{13}{18}\left(MSC:36\right)\)

\(=\dfrac{-28}{36}+\dfrac{15}{36}-\dfrac{26}{36}\)

\(=\dfrac{-13}{36}-\dfrac{26}{36}\)

\(=\dfrac{-39}{36}\)

\(=\dfrac{13}{3}\)

b) \(\dfrac{11}{9}.\dfrac{15}{4}-\dfrac{11}{4}.\dfrac{7}{9}-\dfrac{11}{9}.\dfrac{5}{4}\)

\(=\left(\dfrac{15}{4}-\dfrac{11}{4}-\dfrac{5}{4}\right).\dfrac{11}{9}\)

\(=\left(1-\dfrac{5}{4}\right).\dfrac{11}{9}\)

\(=\left(\dfrac{4}{4}-\dfrac{5}{4}\right).\dfrac{11}{9}\)

\(=\dfrac{-1}{4}.\dfrac{11}{9}\)

\(=\dfrac{-11}{36}\)

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: \(\frac{5}{9}\cdot\frac{7}{13}+\frac{5}{9}\cdot\frac{9}{13}-\frac{5}{9}\cdot\frac{3}{13}\)

\(=\frac{5}{9}\left(\frac{7}{13}+\frac{9}{13}-\frac{5}{13}\right)\)

\(=\frac{5}{9}\)

\(A=\dfrac{1}{3}\cdot\dfrac{4}{5}+\dfrac{1}{3}\cdot\dfrac{6}{5}+\dfrac{2}{3}\\ =\dfrac{1}{3}\cdot\left(\dfrac{4}{5}+\dfrac{6}{5}\right)+\dfrac{2}{3}\\ =\dfrac{1\cdot2}{3}+\dfrac{2}{3}\\ =\dfrac{2}{3}+\dfrac{2}{3}\\ =\dfrac{4}{3}\)

B =\(\dfrac{-5}{6}.\dfrac{4}{19}+\dfrac{-7}{12}.\dfrac{4}{19}\)-\(\dfrac{40}{57}\)

=\(\dfrac{4}{19}.\left[\dfrac{-5}{6}+\dfrac{-7}{12}\right]-\dfrac{40}{57}\)

=\(\dfrac{4}{19}.\left[\dfrac{-10}{12}+\dfrac{-7}{12}\right]-\dfrac{40}{57}\)

=\(\dfrac{4}{19}.\dfrac{-17}{12}-\dfrac{40}{57}\)

=\(\dfrac{-17}{57}\)-\(\dfrac{40}{57}\)

=\(\dfrac{-17}{57}+\dfrac{-40}{57}\)

=\(\dfrac{-57}{57}=-1\)

e. \(\frac{-3}{5}\cdot\frac{5}{7}+\frac{-3}{5}\cdot\frac{3}{7}+\frac{-3}{5}\cdot\frac{6}{7}=-\frac{3}{5}\left(\frac{5}{7}+\frac{3}{7}+\frac{6}{7}\right)=-\frac{3}{5}\cdot2=-\frac{6}{5}\)

f. \(\frac{1}{3}\cdot\frac{4}{5}+\frac{1}{3}\cdot\frac{6}{5}-\frac{4}{5}=\frac{1}{3}\left(\frac{4}{5}+\frac{6}{5}\right)-\frac{4}{5}=\frac{1}{3}\cdot2-\frac{4}{5}=\frac{2}{3}-\frac{4}{5}=-\frac{2}{15}\)

g. \(\frac{4}{19}\cdot\frac{-3}{7}+\frac{-3}{7}\cdot\frac{15}{19}+\frac{5}{7}=-\frac{3}{7}\left(\frac{4}{19}+\frac{15}{19}\right)+\frac{5}{7}=-\frac{3}{7}\cdot1+\frac{5}{7}=-\frac{3}{7}+\frac{5}{7}=\frac{2}{7}\)

h. \(\frac{5}{9}\cdot\frac{7}{13}+\frac{5}{9}\cdot\frac{9}{13}-\frac{5}{9}\cdot\frac{3}{13}=\frac{5}{9}\left(\frac{7}{13}+\frac{9}{13}-\frac{3}{13}\right)=\frac{5}{9}\cdot1=\frac{5}{9}\)

a) \(\left(4,5-2x\right)\cdot1\frac{4}{7}=\frac{11}{7}\)

\(\left(\frac{9}{2}-2x\right)\cdot\frac{11}{7}=\frac{11}{7}\)

\(\left(\frac{9}{2}-2x\right)=\frac{11}{7}\div\frac{11}{7}\)

\(\left(\frac{9}{2}-2x\right)=1\)

\(2x=\frac{9}{2}-1\)

\(x=\frac{7}{2}\div2\)

\(x=\frac{7}{4}\)

b) \(|\frac{3}{4}\cdot x-\frac{1}{2}|-1=\frac{1}{4}\)

\(|\frac{3}{4}\cdot x-\frac{1}{2}|=\frac{1}{4}+1\)

\(|\frac{3}{4}\cdot x|=\frac{5}{4}+\frac{1}{2}\)

\(x=\frac{7}{4}\div\frac{3}{4}\)

\(x=\frac{7}{3}\)

c) \(\frac{1}{4}-|3-x|=-\frac{3}{4}\)

\(|3-x|=\frac{1}{4}-\left(-\frac{3}{4}\right)\)

\(|3-x|=1\)

\(x=3-1\)

\(\Rightarrow x=2\)

d) \(4\cdot\left(x-\frac{6}{7}\right)-\frac{3}{5}=1,4\)

\(4\cdot\left(x-\frac{6}{7}\right)-\frac{3}{5}=\frac{7}{5}\)

\(4\cdot\left(x-\frac{6}{7}\right)=\frac{7}{5}+\frac{3}{5}\)

\(4\cdot\left(x-\frac{6}{7}\right)=2\)

\(\left(x-\frac{6}{7}\right)=2\div4\)

\(x=\frac{1}{2}+\frac{6}{7}\)

\(x=\frac{19}{14}\)

\(\)

k cho mình nhé 

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