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![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(1\right)\dfrac{-7}{12}.\dfrac{11}{8}-\dfrac{37}{8}.\dfrac{7}{12}+\dfrac{1}{2}=-\dfrac{7}{12}.\left(\dfrac{11}{8}+\dfrac{37}{8}\right)+\dfrac{1}{2}=-\dfrac{7}{12}.6+\dfrac{1}{2}=-3.\)
\(\left(2\right)\left(\dfrac{2}{3}-\dfrac{1}{4}-\dfrac{5}{6}\right).\left(-2\right)^2+\dfrac{3}{2}:\dfrac{-15}{4}=\dfrac{-5}{12}.4-\dfrac{2}{5}=\dfrac{-5}{3}-\dfrac{2}{5}=\dfrac{-31}{15}.\)
\(\left(3\right)\dfrac{-2}{5}+\dfrac{3}{10}-\dfrac{3}{5}+\dfrac{7}{10}-\dfrac{3}{2}=1-1-\dfrac{3}{2}=-\dfrac{3}{2}.\)
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.
\(\dfrac{-2}{5}+\dfrac{3}{10}-\dfrac{3}{5}+\dfrac{7}{10}-\dfrac{3}{2}=\dfrac{-4}{10}+\dfrac{3}{10}-\dfrac{6}{10}+\dfrac{7}{10}-\dfrac{15}{10}=\dfrac{-15}{10}=\dfrac{-3}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{-7}{12}.\dfrac{11}{8}-\dfrac{37}{8}.\dfrac{7}{12}+\dfrac{1}{2}=\dfrac{-7}{12}.\left(\dfrac{11}{8}+\dfrac{37}{8}\right)+\dfrac{1}{2}=-\dfrac{7}{12}.6+\dfrac{1}{2}=-\dfrac{7}{2}+\dfrac{1}{2}=-3.\)
\(\left(\dfrac{2}{3}-\dfrac{1}{4}-\dfrac{5}{6}\right).\left(-2\right)^2+\dfrac{3}{2}:\dfrac{-15}{4}=\dfrac{8-3-10}{12}.4+\dfrac{-2}{5}=\dfrac{-5}{12}.4-\dfrac{2}{5}=\dfrac{-5}{3}-\dfrac{2}{5}=-\dfrac{31}{15}.\)
\(\dfrac{-2}{5}+\dfrac{3}{10}-\dfrac{3}{5}+\dfrac{7}{10}-\dfrac{3}{2}=\left(\dfrac{3}{10}+\dfrac{7}{10}\right)+\left(\dfrac{-2}{5}-\dfrac{3}{5}\right)-\dfrac{3}{2}=1-1-\dfrac{3}{2}=\dfrac{-3}{2}.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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: \(=\dfrac{-2}{5}-\dfrac{3}{5}+\dfrac{3}{10}+\dfrac{7}{10}-\dfrac{3}{2}=-\dfrac{3}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\dfrac{7}{30}+\dfrac{\left(-12\right)}{37}+\dfrac{23}{30}+\dfrac{\left(-25\right)}{37}=\left(\dfrac{7}{30}+\dfrac{23}{30}\right)+\left(\dfrac{-12}{37}+\dfrac{-25}{37}\right)=1+\left(-1\right)=0\)
b) \(\dfrac{5}{7}\cdot\dfrac{5}{11}+\dfrac{5}{7}\cdot\dfrac{2}{11}-\dfrac{5}{7}\cdot\dfrac{14}{11}=\dfrac{5}{7}\cdot\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)=\dfrac{5}{7}\cdot\left(-\dfrac{7}{11}\right)=-\dfrac{5}{11}\)
c) \(\dfrac{\left(-5\right)}{7}\cdot\dfrac{3}{13}-\dfrac{5}{7}\cdot\dfrac{10}{13}+1\dfrac{5}{7}=\dfrac{5}{7}\cdot\dfrac{-3}{13}-\dfrac{5}{7}\cdot\dfrac{10}{13}+\dfrac{12}{7}=\dfrac{5}{7}\cdot\left(\dfrac{-3}{13}-\dfrac{10}{13}\right)+\dfrac{12}{7}=\dfrac{5}{7}\cdot\left(-1\right)+\dfrac{12}{7}=\left(-\dfrac{5}{7}\right)+\dfrac{12}{7}=\dfrac{7}{7}=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
c) Ta có: \(\dfrac{3}{5}+\dfrac{-5}{20}+\dfrac{30}{75}+\dfrac{-7}{4}\)
\(=\dfrac{3}{5}+\dfrac{2}{5}+\dfrac{-1}{4}+\dfrac{-7}{4}\)
\(=1-2=-1\)
Giải:
a)-1/12+4/3=-1/12+16/12=15/12=5/4
b)(-4/14-3/15)-(1/5-20/35-(-1)).7
=-17/35-22/35.7
=-17/35-22/5
=-171/35
c)3/5+-5/20+30/75+-7/4
=3/5+-1/4+2/5+-7/4
=(3/5+2/5)+(-1/4+-7/4)
=1+-2
=-1
d)5/6.-12/14+7/13
=-5/7+7/13
=-16/91
e)2/-9-5/-36-1/4
=-1/12-1/4
=-1/3
f)2/23+-5/12+7/18+21/23+-7/12
=(2/23+21/23)+(-5/12+-7/12)+7/18
=1+-1+7/18
=7/18
![](https://rs.olm.vn/images/avt/0.png?1311)
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:\(x=\dfrac{15}{19}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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~~~ Chúc bạn học giỏi ~~~
a) 52 . ( 24 - 12 +1 ) = 52 . ( 16 - 12 + 1 ) = 53
b) 37 . 49 + 37 . 53 - 37 . 2
= 37 . ( 49 + 53 -2 ) = 37 . 100 = 3700
c) 56 : 54 = 56-4 = 52 = 25
d) 79 . 72 : 710 = 711:710 = 7
e) ( 2 + 4 + 6 + ... + 100 ) ( 223 . 335335 - 335 . 223223 )
= ( 2 + 4 + 6 + ... + 100 ) . 0 = 0