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a) \(\frac{8}{40}+\frac{-14}{35}-\frac{12}{60}\)
= \(\frac{1}{5}-\frac{2}{5}-\frac{1}{5}\)
= \(\left(\frac{1}{5}-\frac{1}{5}\right)-\frac{2}{5}\)
= \(-\frac{2}{5}\)
b) 5/7.5/11 + 5/7.5/11 - 5/7.14/11
= 5/7.(5/11 + 5/11 - 14/11)
= 5/7.(-4/11)
= -20/77
c) \(19\frac{5}{8}:\frac{7}{12}-15\frac{1}{4}:\frac{7}{12}\)
= \(\left(19\frac{5}{8}-15\frac{1}{4}\right):\frac{7}{12}\)
= \(\frac{35}{8}:\frac{7}{12}\)
= \(\frac{15}{2}\)
d) 2/5.1/3 - 2/15 : 1/5 + 3/5.1/3
= (2/5 + 3/5).1/3 - 2/15 . 5
= 1.1/3 - 2/3
= 1/3 - 2/3
= -1/3
e) \(\frac{4}{9}.19\frac{1}{3}-\frac{4}{9}.39\frac{1}{3}\)
= \(\frac{4}{9}.\left(19\frac{1}{3}-39\frac{1}{3}\right)\)
= \(\frac{4}{9}.\left(-20\right)\)
= \(\frac{-80}{9}\)
g) \(\frac{81.17-15.81}{81.17-81.15}\)
= 1
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\(=\dfrac{4}{9}\cdot\left(19+\dfrac{1}{3}-39-\dfrac{1}{3}\right)=\dfrac{4}{9}\cdot\left(-20\right)=-\dfrac{80}{9}\)
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a: \(=\dfrac{5}{7}\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)=\dfrac{5}{7}\cdot\dfrac{-7}{11}=-\dfrac{5}{11}\)
b: \(=\dfrac{12}{7}\left(19+\dfrac{5}{8}-15-\dfrac{1}{4}\right)=\dfrac{12}{7}\cdot\left(4+\dfrac{3}{8}\right)\)
\(=\dfrac{12}{7}\cdot\dfrac{35}{8}=\dfrac{3}{2}\cdot5=\dfrac{15}{2}\)
c: \(=\dfrac{2}{15}-\dfrac{2}{15}\cdot5+\dfrac{3}{15}=\dfrac{2}{15}\cdot\left(-4\right)+\dfrac{3}{15}=\dfrac{-8+3}{15}=\dfrac{-5}{15}=-\dfrac{1}{3}\)
d: \(=\dfrac{4}{9}\left(19+\dfrac{1}{3}-39-\dfrac{1}{3}\right)=\dfrac{4}{9}\cdot\left(-20\right)=-\dfrac{80}{9}\)
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\(A=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{37.39}\)
\(2.A=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{37.39}\)
\(2.A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{37}-\frac{1}{39}\)
\(2.A=\frac{1}{3}-\frac{1}{39}\)
\(2.A=\frac{13}{39}-\frac{1}{39}=\frac{12}{39}=\frac{4}{13}\)
\(A=\frac{4}{13}:2=\frac{4}{13}.\frac{1}{2}=\frac{2}{13}\)
\(B=\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{95.96}\)
\(B=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{95}-\frac{1}{96}\)
\(B=\frac{1}{3}-\frac{1}{96}\)
\(B=\frac{32}{96}-\frac{1}{96}=\frac{31}{96}\)
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a) \(\frac{18}{45}-\frac{4}{12}-\frac{6}{9}-\frac{21}{35}=\frac{2}{5}-\frac{1}{3}-\frac{2}{3}-\frac{3}{5}\)
\(=\left(\frac{2}{5}-\frac{3}{5}\right)-\left(\frac{1}{3}+\frac{2}{3}\right)\)
\(=\frac{-1}{3}-1\)
\(=\frac{-4}{3}\)
b) \(\frac{4}{3}+\frac{3}{5}+\frac{7}{3}+\frac{2}{5}+\frac{1}{3}=\left(\frac{4}{3}+\frac{7}{3}+\frac{1}{3}\right)+\left(\frac{3}{5}+\frac{2}{5}\right)\)
\(=4+1=5\)
c) \(\frac{1}{3}.\frac{4}{5}.\frac{1}{3}.\frac{6}{5}=\frac{8}{75}\)
d) \(\frac{6}{7}.\frac{8}{13}+\frac{6}{7}.\frac{9}{13}-\frac{3}{13}.\frac{6}{7}\)
\(=\frac{6}{7}.\left(\frac{8}{13}+\frac{9}{13}-\frac{3}{13}\right)\)
\(=\frac{6}{7}.\frac{14}{13}\)
\(=\frac{12}{13}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2:
a: =>x-35=-23
=>x=12
b: =>|x-8|=13
=>x-8=13 hoặc x-8=-13
=>x=21 hoặc x=-5
Bài 1:
a: =42-98-42+12-12=-98
b: =10x4x3x(-25)=40x(-25)x3=-1000x3=-3000
\(\frac{4}{9}\cdot19\frac{1}{3}-39\frac{1}{3}:\frac{9}{4}\)
\(=\frac{4}{9}\cdot\frac{58}{3}-\frac{118}{3}\cdot\frac{4}{9}\)
\(=\frac{4}{9}\cdot\left(\frac{58}{3}-\frac{118}{3}\right)\)
\(=\frac{4}{9}\cdot\left(-20\right)\)
\(=-\frac{80}{9}\)