\(1\dfrac{4}{25}+\dfrac{5}{21}-0,5+\dfrac{16}{21}\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{27}{23}\) + \(\dfrac{5}{21}\) - \(\dfrac{4}{23}\) + \(0,5\) + \(\dfrac{16}{21}\)
= [ \(\dfrac{27}{23}\) - \(\dfrac{4}{23}\) ] + [ \(\dfrac{5}{21}\) + \(\dfrac{16}{21}\) ] + 0,5
= 1 + 1 + 0,5
= 2 + 0,5
= 2,5
![](https://rs.olm.vn/images/avt/0.png?1311)
a. \(1\dfrac{4}{23}+\dfrac{5}{21}-\dfrac{4}{23}+0,5+\dfrac{16}{21}\)
\(=\left(1\dfrac{4}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{21}+\dfrac{16}{21}\right)+0,5\)
\(=1+1+0,5\)
\(=2,5\)
b. \(\dfrac{3}{7}.19\dfrac{1}{3}-\dfrac{3}{7}.33\dfrac{1}{3}\)
\(=\dfrac{3}{7}.\left(19\dfrac{1}{3}-33\dfrac{1}{3}\right)\)
\(=\dfrac{3}{7}.\left(-14\right)=-6\)
c. \(15\dfrac{1}{4}:\left(-\dfrac{5}{7}\right)-25\dfrac{1}{4}:\left(\dfrac{-5}{7}\right)\)
\(=\left(15\dfrac{1}{4}-25\dfrac{1}{4}\right):\left(-\dfrac{5}{7}\right)\)
\(=-10:\left(-\dfrac{5}{7}\right)\)
\(=14\)
d. \(\left(-\dfrac{2}{3}+\dfrac{3}{7}\right):\dfrac{4}{5}+\left(\dfrac{-1}{3}+\dfrac{4}{7}\right):\dfrac{4}{5}\)
\(=\dfrac{-5}{21}:\dfrac{4}{5}+\dfrac{5}{21}:\dfrac{4}{5}\)
\(=\left(\dfrac{-5}{7}+\dfrac{5}{7}\right):\dfrac{4}{5}\)
\(=0:\dfrac{4}{5}\)
\(=0\)
a,
\(1\dfrac{4}{23}+\dfrac{5}{21}-\dfrac{4}{23}+0,5+\dfrac{16}{21}\)
\(=\left(1\dfrac{4}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{21}+\dfrac{16}{21}\right)+0,5\)
\(=1+1-0,5=1,5\)
b,
\(\dfrac{3}{7}\cdot19\dfrac{1}{3}-\dfrac{3}{7}.33\dfrac{1}{3}\)
\(=\dfrac{3}{7}\left(19\dfrac{1}{3}-33\dfrac{1}{3}\right)=\dfrac{3}{7}.\left(-14\right)=-6\)
c,
\(15\dfrac{1}{4}:\left(-\dfrac{5}{7}\right)-25\dfrac{1}{4}:\left(-\dfrac{5}{7}\right)\)
\(=\left(15\dfrac{1}{4}-25\dfrac{1}{4}\right):\left(-\dfrac{5}{7}\right)=-10:\left(-\dfrac{5}{7}\right)=14\)
d,
\(\left(-\dfrac{2}{3}+\dfrac{3}{7}\right):\dfrac{4}{5}+\left(-\dfrac{1}{3}+\dfrac{4}{7}\right):\dfrac{4}{5}\)
\(=\left(-\dfrac{2}{3}+\dfrac{3}{7}+\dfrac{-1}{3}+\dfrac{4}{7}\right):\dfrac{4}{5}\)
\(=\left[\left(-\dfrac{2}{3}+\dfrac{-1}{3}\right)+\left(\dfrac{3}{7}+\dfrac{4}{7}\right)\right]:\dfrac{4}{5}\)
\(=\left(-1+1\right):\dfrac{4}{5}=0:\dfrac{4}{5}=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)Ta có:\(\left(\dfrac{9}{25}-2,18\right):\left(3\dfrac{4}{5}+0,2\right)=-1,82:4=-\dfrac{91}{200}=-0,455\)
b)Ta có:\(\dfrac{3}{8}.19\dfrac{1}{3}-\dfrac{3}{8}.33\dfrac{1}{3}=\dfrac{3}{8}\left(19\dfrac{1}{3}-33\dfrac{1}{3}\right)=\dfrac{3}{8}.\left(-14\right)=-\dfrac{21}{4}=-5,25\)
c)Ta có:\(1\dfrac{4}{23}+\dfrac{5}{21}-\dfrac{4}{23}+0,5+\dfrac{16}{21}=(1\dfrac{4}{23}-\dfrac{4}{23})+\left(\dfrac{5}{21}+\dfrac{16}{21}\right)+0,5=1+1+0,5=2,5\)Chúc bạn học tốt!!
![](https://rs.olm.vn/images/avt/0.png?1311)
Mình chỉ cho đáp án thôi,sai thì châm chước cho mìn nha!
a)-91 phần200
b)-25phần 4
c)5 phần 2
d)2
e)0
a, ( 0,36-2,18) : ( 3,8 + 0,2)
= -1,82 : 4
=-0,455 hay -91/200
b, 3/8*19/1/3-3/8*33/1/3
=3/8*(19/1/3-33/1/3)
=3/8*(-14)
=-21/4
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(=\left(1+\dfrac{4}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{21}+\dfrac{16}{21}\right)+\dfrac{1}{2}\)
\(=1+1+\dfrac{1}{2}=2+\dfrac{1}{2}=\dfrac{5}{2}\)
b: \(=\left(\dfrac{1}{25}+\dfrac{5}{25}+\dfrac{25}{25}\right):\left(\dfrac{1}{25}-\dfrac{5}{25}-\dfrac{25}{25}\right)\)
\(=\dfrac{31}{25}:\dfrac{-29}{25}=\dfrac{-31}{29}\)
c: \(=\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{\dfrac{4}{9}-\dfrac{4}{7}-\dfrac{4}{11}}+\dfrac{\dfrac{3}{5}-\dfrac{3}{25}-\dfrac{3}{125}-\dfrac{3}{625}}{\dfrac{4}{5}-\dfrac{4}{25}-\dfrac{4}{125}-\dfrac{4}{625}}\)
=1/4+3/4
=1
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải :
a ) \(1\dfrac{4}{23}+\dfrac{5}{21}-\dfrac{4}{23}+0,5+\dfrac{16}{21}\)
\(=\left(1\dfrac{4}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{21}+\dfrac{16}{21}\right)+0,5\)
\(=2,5\)
b ) \(\dfrac{3}{7}.19\dfrac{1}{3}-\dfrac{3}{7}.33\dfrac{1}{3}\)
\(=\dfrac{3}{7}\left(19\dfrac{1}{3}-33\dfrac{1}{3}\right)\)
\(=\dfrac{3}{7}\left(19-33\right)\)
\(=\dfrac{3}{7}\left(-14\right)\)
\(=-6\)
c ) \(9\left(-\dfrac{1}{3}\right)^3+\dfrac{1}{3}\)
\(=9\left(-\dfrac{1}{27}\right)+\dfrac{1}{3}\)
\(=-\dfrac{1}{3}+\dfrac{1}{3}\)
\(=0\)
d ) \(15\dfrac{1}{4}\div\left(-\dfrac{5}{7}\right)-25\dfrac{1}{4}\div\left(-\dfrac{5}{7}\right)\)
\(=\left(15\dfrac{1}{4}-25\dfrac{1}{4}\right)\div\left(-\dfrac{5}{7}\right)\)
\(=-10\left(-\dfrac{7}{5}\right)\)
\(=14\)
\(=\dfrac{29}{25}+1-\dfrac{1}{2}=\dfrac{29}{25}+\dfrac{1}{2}=\dfrac{83}{50}\)