\(\left(1\right)\dfrac{-7}{12}.\dfrac{11}{8}-\dfrac{37}{8}.\dfrac{7}{12}+\dfrac{1}{2}=-\dfrac{7}{12}.\left(\dfrac{11}{8}+\dfrac{37}{8}\right)+\dfrac{1}{2}=-\dfrac{7}{12}.6+\dfrac{1}{2}=-3.\)

\(\left(2\right)\left(\dfrac{2}{3}-\dfrac{1}{4}-\dfrac{5}{6}\right).\left(-2\right)^2+\dfrac{3}{2}:\dfrac{-15}{4}=\dfrac{-5}{12}.4-\dfrac{2}{5}=\dfrac{-5}{3}-\dfrac{2}{5}=\dfrac{-31}{15}.\)

\(\left(3\right)\dfrac{-2}{5}+\dfrac{3}{10}-\dfrac{3}{5}+\dfrac{7}{10}-\dfrac{3}{2}=1-1-\dfrac{3}{2}=-\dfrac{3}{2}.\)

1. \(\dfrac{-7}{12}.\dfrac{11}{8}-\dfrac{37}{8}.\dfrac{7}{12}+\dfrac{1}{2}=\dfrac{-7}{12}\left(\dfrac{11}{8}+\dfrac{37}{8}\right)+\dfrac{1}{2}=\dfrac{-7}{12}.\dfrac{6}{1}+\dfrac{1}{2}=\dfrac{-7}{2}+\dfrac{1}{2}=\dfrac{-6}{2}=-3\)2.
\(\left(\dfrac{2}{3}-\dfrac{1}{4}-\dfrac{5}{6}\right).\left(-2\right)^2+\dfrac{3}{2}:\dfrac{-15}{4}=\dfrac{-5}{12}.4+\dfrac{-2}{5}=\dfrac{-5}{3}+\dfrac{-2}{5}=\dfrac{-31}{15}\)
3.
\(\dfrac{-2}{5}+\dfrac{3}{10}-\dfrac{3}{5}+\dfrac{7}{10}-\dfrac{3}{2}=\dfrac{-4}{10}+\dfrac{3}{10}-\dfrac{6}{10}+\dfrac{7}{10}-\dfrac{15}{10}=\dfrac{-15}{10}=\dfrac{-3}{2}\)

\(\dfrac{-7}{12}.\dfrac{11}{8}-\dfrac{37}{8}.\dfrac{7}{12}+\dfrac{1}{2}=\dfrac{-7}{12}.\left(\dfrac{11}{8}+\dfrac{37}{8}\right)+\dfrac{1}{2}=-\dfrac{7}{12}.6+\dfrac{1}{2}=-\dfrac{7}{2}+\dfrac{1}{2}=-3.\)

\(\left(\dfrac{2}{3}-\dfrac{1}{4}-\dfrac{5}{6}\right).\left(-2\right)^2+\dfrac{3}{2}:\dfrac{-15}{4}=\dfrac{8-3-10}{12}.4+\dfrac{-2}{5}=\dfrac{-5}{12}.4-\dfrac{2}{5}=\dfrac{-5}{3}-\dfrac{2}{5}=-\dfrac{31}{15}.\)

\(\dfrac{-2}{5}+\dfrac{3}{10}-\dfrac{3}{5}+\dfrac{7}{10}-\dfrac{3}{2}=\left(\dfrac{3}{10}+\dfrac{7}{10}\right)+\left(\dfrac{-2}{5}-\dfrac{3}{5}\right)-\dfrac{3}{2}=1-1-\dfrac{3}{2}=\dfrac{-3}{2}.\)

1: \(=\dfrac{7}{12}\left(-\dfrac{11}{8}+\dfrac{37}{8}\right)+\dfrac{1}{2}=\dfrac{7}{12}\cdot\dfrac{26}{8}+\dfrac{1}{2}=\dfrac{115}{48}\)

2: \(=\dfrac{8-3-10}{12}\cdot4+\dfrac{3}{2}\cdot\dfrac{-4}{15}\)

\(=\dfrac{-5}{3}+\dfrac{-12}{30}=\dfrac{-5}{3}+\dfrac{-2}{5}=\dfrac{-25-6}{15}=-\dfrac{31}{15}\)

3: \(=\dfrac{-2}{5}-\dfrac{3}{5}+\dfrac{3}{10}+\dfrac{7}{10}-\dfrac{3}{2}=-\dfrac{3}{2}\)

a)\(\left(\frac{3}{4}+\frac{-7}{2}\right).\left(\frac{10}{11}+\frac{2}{22}\right)\)

\(\left(\frac{3}{4}+\frac{-14}{4}\right).\left(\frac{20}{22}+\frac{2}{22}\right)\)

\(\frac{-11}{4}.1=\frac{-11}{4}\)

câu b mk hiểu đề

0,75+7/12 chứ không phải = đâu

Bài 3 là hỗn số hả em?

Bài 1: 

a) Ta có: \(\dfrac{2}{5}\cdot x+\dfrac{1}{3}=\dfrac{1}{5}\)

\(\Leftrightarrow\dfrac{2}{5}\cdot x=\dfrac{1}{5}-\dfrac{1}{3}=\dfrac{-2}{15}\)

\(\Leftrightarrow x=\dfrac{-2}{15}:\dfrac{2}{5}=\dfrac{-2}{15}\cdot\dfrac{5}{2}\)

hay \(x=-\dfrac{1}{3}\)

Vậy: \(x=-\dfrac{1}{3}\)

b) Ta có: \(\dfrac{1}{5}+\dfrac{5}{3}:x=\dfrac{1}{2}\)

\(\Leftrightarrow\dfrac{5}{3}:x=\dfrac{1}{2}-\dfrac{1}{5}=\dfrac{3}{10}\)

\(\Leftrightarrow x=\dfrac{5}{3}:\dfrac{3}{10}=\dfrac{5}{3}\cdot\dfrac{10}{3}\)

hay \(x=\dfrac{50}{9}\)

Vậy: \(x=\dfrac{50}{9}\)

c) Ta có: \(\dfrac{4}{9}-\dfrac{5}{3}\cdot x=-2\)

\(\Leftrightarrow\dfrac{5}{3}x=\dfrac{4}{9}+2=\dfrac{22}{9}\)

\(\Leftrightarrow x=\dfrac{22}{9}:\dfrac{5}{3}=\dfrac{22}{9}\cdot\dfrac{3}{5}\)

hay \(x=\dfrac{22}{15}\)

Vậy: \(x=\dfrac{22}{15}\)

d) Ta có: \(\dfrac{5}{7}:x-3=\dfrac{-2}{7}\)

\(\Leftrightarrow\dfrac{5}{7}:x=\dfrac{-2}{7}+3=\dfrac{19}{21}\)

\(\Leftrightarrow x=\dfrac{5}{7}:\dfrac{19}{21}=\dfrac{5}{7}\cdot\dfrac{21}{19}\)

hay \(x=\dfrac{15}{19}\)

Vậy:\(x=\dfrac{15}{19}\)

a) \(\left(\dfrac{2}{3}\right)^3-\left(\dfrac{3}{4}\right)^2.\left(-1\right)^5\)

\(=\dfrac{8}{27}-\dfrac{9}{16}.\left(-1\right)\)

\(=\dfrac{8}{27}-\left(-\dfrac{9}{16}\right)\)

\(=\dfrac{8}{27}+\dfrac{9}{16}\)

\(=\dfrac{128}{432}+\dfrac{243}{432}\)

\(=\dfrac{371}{432}\)

b) \(12:\left(\dfrac{3}{4}-\dfrac{5}{6}\right)^2\)

\(=12:\left(\dfrac{9}{12}-\dfrac{10}{12}\right)^2\)

\(=12:\left(\dfrac{-1}{12}\right)^2\)

\(=12:\dfrac{1}{144}\)

\(=12.144\)

\(=1728\)

c) \(\dfrac{7}{22}:\dfrac{3}{11}+\dfrac{7}{22}:\dfrac{4}{11}\)

\(=\dfrac{7}{22}:\left(\dfrac{3}{11}+\dfrac{4}{11}\right)\)

\(=\dfrac{7}{22}:\dfrac{7}{11}\)

\(=\dfrac{7}{22}.\dfrac{11}{7}\)

\(=\dfrac{1}{2}\)

d) \(\dfrac{12}{35}.\left(\dfrac{7}{4}+\dfrac{13}{4}\right)-\dfrac{1}{3}\)

\(=\dfrac{12}{35}.5-\dfrac{1}{3}\)

\(=\dfrac{12}{7}-\dfrac{1}{3}\)

\(=\dfrac{36}{21}-\dfrac{7}{21}\)

\(=\dfrac{29}{21}\)

a,

A=1−3−5−7−9−...−97−99a)A=1−3−5−7−9−...−97−99 

=1−(3+5+7+...+99)=1−(3+5+7+...+99)

=1−(99+3).[(99−3):2+1]2=1−(99+3).[(99−3):2+1]2
=1−2499=−2498=1−2499=−2498

b)B=1+3−5−7+9+...+97−99b)B=1+3−5−7+9+...+97−99
=(−8)+(−8)+(−8)+...+(−8)+97−99=(−8)+(−8)+(−8)+...+(−8)+97−99
=(−8).12+(−2)=−98=(−8).12+(−2)=−98

c)C=1−3−5+7+9−11−13+15+...+97−99c)C=1−3−5+7+9−11−13+15+...+97−99
=0+0+0+0+0+...+0−99=0+0+0+0+0+...+0−99
=−99