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![](https://rs.olm.vn/images/avt/0.png?1311)
Giải:
a) \(2\dfrac{17}{20}-1\dfrac{15}{11}+6\dfrac{9}{20}:3\)
\(=\dfrac{57}{20}-\dfrac{26}{11}+\dfrac{129}{20}:3\)
\(=\dfrac{107}{220}+\dfrac{43}{20}\)
\(=\dfrac{29}{11}\)
b) \(4\dfrac{3}{7}:\left(\dfrac{7}{5}.4\dfrac{3}{7}\right)\)
\(=\dfrac{31}{7}:\left(\dfrac{7}{5}.\dfrac{31}{7}\right)\)
\(=\dfrac{31}{7}:\dfrac{31}{5}\)
\(=\dfrac{5}{7}\)
c) \(\left(3\dfrac{2}{9}.\dfrac{15}{23}.1\dfrac{7}{29}\right):\dfrac{5}{23}\)
\(=\left(\dfrac{29}{9}.\dfrac{15}{23}.\dfrac{36}{29}\right):\dfrac{5}{23}\)
\(=\dfrac{60}{23}:\dfrac{5}{23}\)
\(=12\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a).\(\dfrac{x}{15}\)=\(\dfrac{4}{3}\)\(\Leftrightarrow\)x.3=15.4\(\Leftrightarrow\)3x=60\(\Leftrightarrow\)x=20
b)x:\(\dfrac{4}{3}\)=\(\dfrac{5}{6}\)\(\Leftrightarrow\)\(\dfrac{3x}{4}=\dfrac{5}{6}\)\(\Leftrightarrow\)3x.6=4.5\(\Leftrightarrow\)18x=20\(\Leftrightarrow\)\(\dfrac{10}{9}\)
c)\(\dfrac{7}{2}\)-x=\(\dfrac{5}{6}\)\(\Leftrightarrow\)\(\dfrac{7}{2}-\dfrac{2x}{2}\)=\(\dfrac{5}{6}\)\(\Leftrightarrow\)\(\dfrac{7-2x}{2}\)=\(\dfrac{5}{6}\)\(\Leftrightarrow\)(7-2x).6=2.5\(\Leftrightarrow\)42-12x=10
\(\Leftrightarrow\)-12x=-32\(\Leftrightarrow\)x=\(\dfrac{8}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1 :
a, \(\frac{3}{4}:x=\frac{5}{12}\)
\(x=\frac{3}{4}:\frac{5}{12}\)
\(x=\frac{9}{5}\)
b, \(x-\frac{1}{2}=\frac{3}{4}:\frac{3}{2}\)
\(x-\frac{1}{2}=\frac{1}{2}\)
\(x=\frac{1}{2}+\frac{1}{2}\)
\(x=1\)
c, \(1\frac{1}{2}x-\frac{1}{2}=\frac{3}{4}\)
\(\frac{3}{2}x-\frac{1}{2}=\frac{3}{4}\)
\(\frac{3}{2}x=\frac{3}{4}+\frac{1}{2}\)
\(\frac{3}{2}x=\frac{5}{4}\)
\(x=\frac{5}{4}:\frac{3}{2}\)
\(x=\frac{5}{6}\)
Bài 2 :
\(A=\frac{-3}{5}+\left(\frac{-2}{5}-99\right)\)
\(A=\frac{-3}{5}+\frac{-2}{5}-99\)
\(A=\left(-1\right)-99\)
\(A=-100\)
\(B=\left(7\frac{2}{3}+2\frac{3}{5}\right)-6\frac{2}{3}\)
\(B=\left(\frac{23}{3}+\frac{13}{5}\right)-\frac{20}{3}\)
\(B=\frac{23}{3}+\frac{13}{5}-\frac{20}{3}\)
\(B=\left(\frac{23}{3}-\frac{20}{3}\right)+\frac{13}{5}\)
\(B=1+\frac{13}{5}\)
\(B=\frac{18}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\dfrac{-3}{20}\) + \(\dfrac{-7}{4}\) =\(\dfrac{-3}{20}\) + \(\dfrac{-35}{20}\) = -2
b) 6 và \(\dfrac{2}{3}\) - 4 và \(\dfrac{2}{3}\) = 2
c) \(\dfrac{-3}{10}\) + \(\dfrac{7}{12}\) = \(\dfrac{-18}{60}\) + \(\dfrac{35}{60}\) =\(\dfrac{17}{60}\)
d) \(\dfrac{35}{-9}\) . \(\dfrac{81}{7}\) = \(\dfrac{-35}{9}\) . \(\dfrac{81}{7}\) = 45
e) \(\dfrac{-2}{5}\) - \(\dfrac{-3}{4}\) = \(\dfrac{-8}{20}\) - \(\dfrac{-15}{20}\) = \(\dfrac{-8}{20}\) + \(\dfrac{15}{20}\) =\(\dfrac{7}{20}\)
f) \(\dfrac{5}{23}\) . \(\dfrac{7}{26}\) + \(\dfrac{5}{23}\) .\(\dfrac{9}{26}\) = \(\dfrac{5}{23}\) . ( \(\dfrac{7}{26}\) + \(\dfrac{9}{26}\) )= \(\dfrac{5}{23}\) . \(\dfrac{8}{13}\) = \(\dfrac{40}{299}\)
g) \(\dfrac{-3}{12}\) : \(\dfrac{4}{15}\) =\(\dfrac{-3}{12}\) . \(\dfrac{15}{4}\) =\(\dfrac{-5}{8}\)
h) 1 và \(\dfrac{1}{6}\) - 3 và \(\dfrac{1}{3}\) =\(\dfrac{7}{6}\) -\(\dfrac{10}{3}\) = \(\dfrac{-13}{6}\)
i) \(\dfrac{-2}{5}\) . (-3) + \(\dfrac{3}{8}\) . \(\dfrac{4}{-10}\) =(\(\dfrac{-2}{5}\) .\(\dfrac{-4}{10}\)) + [(-3) . \(\dfrac{3}{8}\)
= \(\dfrac{4}{25}\) + \(\dfrac{-9}{8}\) = \(\dfrac{32}{200}\) + \(\dfrac{-225}{200}\) = \(\dfrac{-193}{200}\)
j) \(\dfrac{-13}{17}\) + (\(\dfrac{13}{-21}\) + \(\dfrac{-4}{17}\) )
= ( \(\dfrac{-13}{17}\) + \(\dfrac{-4}{17}\) )+\(\dfrac{-13}{21}\)
= -1+\(\dfrac{-13}{21}\)
= \(\dfrac{-21}{21}\) + \(\dfrac{-13}{21}\) = \(\dfrac{-34}{21}\)
Khôi nguyễn
![](https://rs.olm.vn/images/avt/0.png?1311)
4:
a: =4/15-2,9+11/15=1-2,9=-1,9
b: \(=-36,75+3,7-63,25+6,3=10-100=-90\)
c: \(=6,5+3,5-\dfrac{10}{17}-\dfrac{7}{17}=10-1=9\)
d: \(=\dfrac{13}{25}\left(-39,1-60,9\right)=\dfrac{13}{25}\left(-100\right)=-52\)
e: =-5/12-7/12-3,7-6,3=-1-10=-11
f: =2,8(-6/13-7/13)-7,2=-2,8-7,2=-10
![](https://rs.olm.vn/images/avt/0.png?1311)
a: =>4/3x=7/9-4/9=1/3
=>x=1/4
b: =>5/2-x=9/14:(-4/7)=-9/8
=>x=5/2+9/8=29/8
c: =>3x+3/4=8/3
=>3x=23/12
hay x=23/36
d: =>-5/6-x=7/12-4/12=3/12=1/4
=>x=-5/6-1/4=-10/12-3/12=-13/12
![](https://rs.olm.vn/images/avt/0.png?1311)
1: 2*2*2*2*2*2=2^6
2: x*x*x*x=x^4
3: =7^4*6^4=42^4
4: =5^2*3^2*4^2=60^2
5: =2*3^2*5^5
6: =10^3*10^3=10^6
a) \(\frac{3}{4}\)+ \(\frac{4}{3}\):\(\frac{2}{7}\)- \(\frac{4}{3}\)
=\(\frac{3}{4}\) +\(\frac{4}{3}\) .\(\frac{7}{2}\) -\(\frac{4}{3}\)
=\(\frac{3}{4}\) + \(\frac{14}{3}\) -\(\frac{4}{3}\)
=\(\frac{9}{12}\) +\(\frac{56}{12}\)
=\(\frac{56}{12}\) - \(\frac{4}{3}\)
=\(\frac{56}{12}\)- \(\frac{16}{12}\)
=\(\frac{10}{3}\)
b) \(\frac{5}{2}\) - [\(\frac{1}{2}\) + (\(\frac{-2}{3}\)) .\(\frac{6}{5}\)].\(\frac{2}{5}\)
= \(\frac{5}{2}\) - [\(\frac{1}{2}\) + (\(\frac{-4}{5}\))].\(\frac{2}{5}\)
= \(\frac{5}{2}\) - (\(\frac{5}{10}\) +\(\frac{-8}{10}\)).\(\frac{2}{5}\)
=\(\frac{5}{2}\)-\(\frac{3}{10}\).\(\frac{2}{5}\)
=\(\frac{5}{2}\)-\(\frac{3}{25}\)
= \(\frac{125}{50}\)-\(\frac{6}{50}\)
= \(\frac{119}{50}\)
(Mình không chắc chắn lắm!^^)