=1+4/23-4/23+5/21+16/21+1/2

=1+1+1/2

=5/2

\(\dfrac{27}{23}\) + \(\dfrac{5}{21}\) - \(\dfrac{4}{23}\) + \(0,5\) + \(\dfrac{16}{21}\)

= [ \(\dfrac{27}{23}\) - \(\dfrac{4}{23}\) ] + [ \(\dfrac{5}{21}\) + \(\dfrac{16}{21}\) ] + 0,5

=  1 + 1 + 0,5 

=  2 + 0,5

=  2,5

`21/25-5/21-[-4]/25-16/21+1/2`

`=(21/25-[-4]/25)-(5/21+16/21)+1/2`

`=25/25-21/21+1/2`

`=1-1+1/2=1/2`

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,

\(\left(-\dfrac{2}{3}+\dfrac{3}{7}\right):\dfrac{4}{5}+\left(-\dfrac{1}{3}+\dfrac{4}{7}\right):\dfrac{4}{5}\)

\(=\left(-\dfrac{2}{3}+\dfrac{3}{7}+\dfrac{-1}{3}+\dfrac{4}{7}\right):\dfrac{4}{5}\)

\(=\left[\left(-\dfrac{2}{3}+\dfrac{-1}{3}\right)+\left(\dfrac{3}{7}+\dfrac{4}{7}\right)\right]:\dfrac{4}{5}\)

\(=\left(-1+1\right):\dfrac{4}{5}=0:\dfrac{4}{5}=0\)

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!!

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

sorry

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

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 ) \(15\dfrac{1}{4}\div\left(-\dfrac{5}{7}\right)-25\dfrac{1}{4}\div\left(-\dfrac{5}{7}\right)\)

\(=\left(15\dfrac{1}{4}-25\dfrac{1}{4}\right)\div\left(-\dfrac{5}{7}\right)\)

\(=-10\left(-\dfrac{7}{5}\right)\)

\(=14\)