bài1  

a) \(\dfrac{7}{6}-\dfrac{13}{12}+\dfrac{3}{4}\) 

=\(\dfrac{14}{12}-\dfrac{13}{12}+\dfrac{9}{12}\) 

=\(\dfrac{1}{12}+\dfrac{9}{12}\) 

=\(\dfrac{10}{12}=\dfrac{5}{6}\)

bài 1 

b)\(1\dfrac{1}{2}.(\dfrac{-4}{5})\) + \(\dfrac{3}{10}\) 

= \(\dfrac{3}{2}.\left(-\dfrac{4}{5}\right)+\dfrac{3}{10}\) 

= \(-\dfrac{6}{5}+\dfrac{3}{10}\) 

=\(-\dfrac{12}{10}+\dfrac{3}{10}\) 

=\(-\dfrac{9}{10}\) 

=50/25=2

55/25 = 11/5

55/25

a) Ta có: \(\dfrac{x}{3}=\dfrac{7}{25}+\dfrac{-1}{5}\)

\(\Leftrightarrow\dfrac{x}{3}=\dfrac{7}{25}+\dfrac{-5}{25}=\dfrac{2}{25}\)

hay \(x=\dfrac{6}{25}\)

Vậy: \(x=\dfrac{6}{25}\)

b) Ta có: \(\dfrac{4}{9}+\dfrac{x}{5}=\dfrac{5}{11}\)

\(\Leftrightarrow\dfrac{x}{5}=\dfrac{5}{11}-\dfrac{4}{9}=\dfrac{45}{99}-\dfrac{44}{99}=\dfrac{1}{99}\)

hay \(x=\dfrac{5}{99}\)

Vậy: \(x=\dfrac{5}{99}\)

c) Ta có: \(\dfrac{-5}{9}+\dfrac{x}{10}=\dfrac{1}{3}\)

\(\Leftrightarrow\dfrac{x}{10}=\dfrac{1}{3}+\dfrac{5}{9}=\dfrac{3}{9}+\dfrac{5}{9}=\dfrac{8}{9}\)

hay \(x=\dfrac{80}{9}\)

Vậy: \(x=\dfrac{80}{9}\)

a: \(=\dfrac{14-2+9}{32}\cdot\dfrac{4}{5}=\dfrac{21}{5}\cdot\dfrac{1}{8}=\dfrac{21}{40}\)

b: \(=10+\dfrac{2}{9}+2+\dfrac{3}{5}+6+\dfrac{2}{9}=18+\dfrac{47}{45}=\dfrac{857}{45}\)

c: \(=\dfrac{3}{10}-\dfrac{12}{5}+\dfrac{1}{10}=\dfrac{4}{10}-\dfrac{12}{5}=\dfrac{2}{5}-\dfrac{12}{5}=-2\)

d: \(=\dfrac{-25}{30}\left(\dfrac{37}{44}+\dfrac{13}{44}-\dfrac{6}{44}\right)=\dfrac{-25}{30}\cdot1=-\dfrac{5}{6}\)

a) \(1-\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{6}{6}-\dfrac{3}{6}+\dfrac{2}{6}=\dfrac{6-3+2}{6}=\dfrac{1}{6}\)

\(b.\) \(\dfrac{2}{5}+\dfrac{3}{5}:\dfrac{9}{10}=\dfrac{2}{5}+\dfrac{3}{5}.\dfrac{10}{9}=\dfrac{2}{5}+\dfrac{2}{3}=\dfrac{6}{15}+\dfrac{10}{15}=\dfrac{6+10}{15}=\dfrac{16}{15}\)

\(c.\) \(\dfrac{7}{11}.\dfrac{3}{4}+\dfrac{7}{11}.\dfrac{1}{4}+\dfrac{4}{11}=\dfrac{21}{44}+\dfrac{7}{44}+\dfrac{4}{11}=\dfrac{21}{44}+\dfrac{7}{44}+\dfrac{16}{44}=\dfrac{21+7+16}{44}=\dfrac{44}{44}=1\)

a/\(1-\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{6}{6}-\dfrac{3}{6}+\dfrac{2}{6}=\dfrac{5}{6}\)

b/\(\dfrac{2}{5}+\dfrac{3}{5}:\dfrac{9}{10}=\dfrac{2}{5}+\dfrac{3}{5}.\dfrac{10}{9}=\dfrac{2}{5}+\dfrac{2}{3}=\dfrac{6}{15}+\dfrac{10}{15}=\dfrac{16}{15}\)

d) \(\left(\dfrac{3}{4}+0,5+25\%\right).2\dfrac{2}{3}=\left(\dfrac{3}{4}+\dfrac{1}{2}+\dfrac{1}{4}\right).\dfrac{8}{3}=\left(\dfrac{3}{4}+\dfrac{2}{4}+\dfrac{1}{4}\right).\dfrac{8}{3}=\dfrac{3}{2}.\dfrac{8}{3}=4\)

43/14

1/3

71/15

43/14

1/3

71/15

\(\dfrac{43}{14}\)

\(\dfrac{1}{3}\)

\(\dfrac{71}{15}\)

o: \(\dfrac{\left(-1\right)^6\cdot3^5\cdot4^3}{9^2\cdot2^5}=\dfrac{3^5\cdot2^6}{2^5\cdot3^4}=\dfrac{3^5}{3^4}\cdot\dfrac{2^6}{2^5}=3\cdot2=6\)

s: \(\dfrac{\dfrac{2}{7}+\dfrac{2}{5}+\dfrac{2}{17}-\dfrac{2}{25}}{\dfrac{3}{14}+\dfrac{3}{10}+\dfrac{3}{34}-\dfrac{3}{50}}\)

\(=\dfrac{2\left(\dfrac{1}{7}+\dfrac{1}{5}+\dfrac{1}{17}-\dfrac{1}{25}\right)}{\dfrac{3}{2}\left(\dfrac{1}{7}+\dfrac{1}{5}+\dfrac{1}{17}-\dfrac{1}{25}\right)}\)

\(=2:\dfrac{3}{2}=\dfrac{4}{3}\)

t: \(\sqrt{\dfrac{4}{9}}-\dfrac{1}{2}:\left|-\dfrac{2}{3}\right|\)

\(=\dfrac{2}{3}-\dfrac{1}{2}:\dfrac{2}{3}\)

\(=\dfrac{2}{3}-\dfrac{3}{4}=\dfrac{8-9}{12}=-\dfrac{1}{12}\)

a)\(-160\)

b)\(-7000\)

c)\(-\dfrac{1}{12}\)

d)\(-\dfrac{3}{2}\)

a) \(\dfrac{2}{9}+\dfrac{5}{9}=\dfrac{7}{9};\dfrac{5}{9}+\dfrac{2}{9}=\dfrac{7}{9}\)

Vậy: \(\dfrac{2}{9}+\dfrac{5}{9}=\dfrac{5}{9}+\dfrac{2}{9}\)

b) \(\dfrac{3}{25}+\dfrac{4}{25}+\dfrac{7}{25}=\dfrac{14}{25};\dfrac{3}{25}+\dfrac{7}{25}+\dfrac{4}{25}=\dfrac{14}{25}\)

Vậy: \(\dfrac{3}{25}+\dfrac{4}{25}+\dfrac{7}{25}=\dfrac{3}{25}+\dfrac{7}{25}+\dfrac{4}{25}\)