\(\left(19\frac{5}{8}:\frac{7}{12}-13\frac{1}{4}:\frac{7}{12}\right).\frac{4}{5}\)

\(=\left(\frac{157}{8}:\frac{7}{12}-\frac{53}{4}:\frac{7}{12}\right).\frac{4}{5}\)

\(=\left[\left(\frac{157}{8}-\frac{53}{4}\right):\frac{7}{12}\right].\frac{4}{5}\)

\(=\left[\frac{51}{8}:\frac{7}{12}\right].\frac{4}{5}\)

\(=\frac{153}{14}.\frac{4}{5}\)

\(=\frac{306}{35}\)

\(\left(\frac{-2}{5}+\frac{3}{7}\right)-\left(\frac{4}{9}+\frac{12}{20}-\frac{13}{35}\right)+\frac{7}{35}\)

\(=\frac{1}{35}-\frac{212}{315}+\frac{7}{35}\)

\(=\frac{1}{35}+\frac{-212}{315}+\frac{7}{35}\)

\(=\frac{9}{315}+\frac{-212}{315}+\frac{63}{315}\)

\(=\frac{-140}{315}=\frac{-4}{9}\)

\(\left(4\frac{5}{57}-3\frac{4}{51}+8\frac{13}{29}\right)-\left(3\frac{5}{57}-6\frac{16}{29}\right)\)

\(=\left(\frac{233}{57}-\frac{157}{51}+\frac{245}{29}\right)-\left(\frac{176}{57}-\frac{190}{29}\right)\)

\(=\frac{1941}{323}-\frac{-5726}{1653}\)

\(=\frac{14011}{1479}\)

3: \(=\dfrac{13}{5}\left(-\dfrac{3}{14}+\dfrac{2}{5}+\dfrac{-11}{14}+\dfrac{3}{5}\right)\)

=0

a: =(5/7+2/7)+(4/3+5/3)=3+1=4

b: =(17/12+7/12)+(29/7-8/7)

=2+3=5

c: =(2/5+3/5)+(6/9+1/3)+(7/4+1/4)

=1+2+1

=4

d: =(1/3+2/3)+(13/17+4/17)+(29/11+4/11)

=1+1+3=5

Giải:

a) \(\left(9\dfrac{4}{9}+5\dfrac{2}{3}\right)-5\dfrac{1}{2}\) 

\(=\left(\dfrac{85}{9}+\dfrac{17}{3}\right)-\dfrac{11}{2}\) 

\(=\dfrac{136}{9}-\dfrac{11}{2}\) 

\(=\dfrac{173}{18}\) 

b) \(\dfrac{13}{9}.\dfrac{15}{4}-\dfrac{13}{9}.\dfrac{7}{4}-\dfrac{13}{9}.\dfrac{5}{4}\) 

\(=\dfrac{13}{9}.\left(\dfrac{15}{4}-\dfrac{7}{4}-\dfrac{5}{4}\right)\) 

\(=\dfrac{13}{9}.\dfrac{3}{4}\) 

\(=\dfrac{13}{12}\) 

c) \(\dfrac{2}{3}+\dfrac{5}{8}-\dfrac{-1}{3}+0,375\) 

\(=\left(\dfrac{2}{3}-\dfrac{-1}{3}\right)+\left(\dfrac{5}{8}+\dfrac{3}{8}\right)\) 

\(=1+1\) 

\(=2\)

d) \(75\%-3\dfrac{1}{2}+1,5:\dfrac{10}{7}\) 

\(=\dfrac{3}{4}+\dfrac{7}{2}+\dfrac{3}{2}:\dfrac{10}{7}\) 

\(=\dfrac{3}{4}+\dfrac{7}{2}+\dfrac{21}{20}\) 

\(=\dfrac{53}{10}\) 

e) \(1\dfrac{13}{15}.\left(0,5\right)^2.3+\left(\dfrac{8}{15}-1\dfrac{19}{60}\right):1\dfrac{23}{24}\) 

\(=\dfrac{28}{15}.\dfrac{1}{4}.3+\left(\dfrac{8}{15}-\dfrac{79}{60}\right):\dfrac{47}{24}\) 

\(=\dfrac{7}{5}+\dfrac{-47}{60}:\dfrac{47}{24}\) 

\(=\dfrac{7}{5}+\dfrac{-2}{5}\) 

\(=1\)

a) = 17/19 - 17/19 + 27/35 + 35/35 = 0 + 62/35

b) = 1/3 x 4/5 + 1/3 x6/5 + 1/3 x 2 = 1/3(4/5 + 6/5 + 2) = 1/3 x 4 = = 4/3

c) 4/7 x 2/9 + 4/7 x 7/9 + 2/3 = 4/7 x (2/9 + 7/9) + 2/3 = 4/7 x 1 + 2/3 = 26/21

A) 17/19 - 17/19 + 27/35 + 35/35 = 0 + 62/35

B) 1/3 x 4/5 + 1/3 x 6/5 + 1/3 x 2 = 1/3 x(4/5 + 6/5 x 2 ) = 1/3 x 4 = 4/3

c) TƯƠNG TỰ CÂU A VÀ B

* HOKTOT*

NHA

Bài 1: 

Ta có: \(x-35\%\cdot x=\dfrac{1}{25}\)

\(\Leftrightarrow65\%\cdot x=\dfrac{1}{25}\)

\(\Leftrightarrow x=\dfrac{1}{25}:\dfrac{13}{20}=\dfrac{1}{25}\cdot\dfrac{20}{13}=\dfrac{4}{65}\)

Vậy: \(x=\dfrac{4}{65}\)

Bài 2: 

a) Ta có: \(17\dfrac{2}{31}-\left(\dfrac{15}{17}+6\dfrac{2}{31}\right)\)

\(=17\dfrac{2}{31}-\dfrac{15}{17}-6\dfrac{2}{31}\)

\(=11+\dfrac{2}{31}-\dfrac{15}{17}\)

\(=\dfrac{5366}{527}\)

Np 3  và Sl 2 có cùng chung điểm Suy ra

a: \(\dfrac{3}{5}-\dfrac{7}{4}=\dfrac{12}{20}-\dfrac{35}{20}=\dfrac{-23}{20}\)

b: \(\dfrac{4}{5}:\dfrac{-8}{15}=\dfrac{-4}{5}\cdot\dfrac{15}{8}=\dfrac{-60}{40}=\dfrac{-3}{2}\)

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

d: \(=\dfrac{-3}{7}\left(\dfrac{4}{19}+\dfrac{15}{19}\right)+\dfrac{5}{7}=\dfrac{2}{7}\)