\(\dfrac{15}{11}\)+\(\dfrac{8}{13}\)-\(\...">
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9 tháng 12 2017

\(\dfrac{15}{11}+\dfrac{8}{13}-\dfrac{4}{11}+\dfrac{18}{13}\)

\(=\left(\dfrac{15}{11}-\dfrac{4}{11}\right)+\left(\dfrac{8}{13}+\dfrac{18}{13}\right)\)

\(=1+2\)

\(=3\)

20 tháng 12 2017

\(\dfrac{15}{11}+\dfrac{8}{13}-\dfrac{4}{11}+\dfrac{18}{13}=\left(\dfrac{15}{11}-\dfrac{4}{11}\right)+\left(\dfrac{8}{13}+\dfrac{18}{13}\right)=\dfrac{11}{11}+\dfrac{26}{13}=1+2=3\)

29 tháng 7 2018

A)0,25:(10,3-9,8)-3/4

=1/4:(103/10-49/5)-3/4

=1/4:1/2-3/4

=1/2-3/4

=2/4-3/4

=-1/4

B)-5/9.13/28-13/28.4/9

=-5/9-4/9.13/28

=-1.13/28

=-13/28

c)6/7+5/8:5-3/16

=6/7+1/8-3/16

=55/56-3/16

=89/112

d)-5/7.2/11+-5/7.9/11+1/5/7

=-5/7.(2/11+9/11)+12/7

=-5/7.1+12/7

=-5/7+12/7

=1

e)-7/12-8/15+11/20

=-67/60+11/20

=-17/30

f)-17/25.20/33+-17/25.13/33+-3/25

=-17/25.(20/33+13/33)-3/25

=-17/25.1-3/25

=-17/25-3/25

=-4/5

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4 tháng 8 2018

Ta có : a, 25/7 + 13/21 - 11/7 + 17/21 + 1/3 .

= ( 25/7 - 11/7 ) + ( 13/21 + 17/21 + 1/3 ) .

= 2 + ( 20/21 + 7/21 ) .

= 2 + 9/7 .

= 23/7 .

b, ( 1/3 + 12/67 + 13/41 ) - ( 79/67 - 28/41 ) .

= 1/3 + 12/67 + 13/41 - 79/67 + 28/41 .

= 1/3 + ( 12/67 - 79/67 ) + ( 13/41 + 28/41 ) .

= 1/3 - 1 + 1 .

= 1/3 .

c, ( 11/4 . -5/9 - 4/9 . 11/4 ) . 8/33 .

= [ 11/4 . ( -5/9 - 4/9 ) ] . 8/33 .

= [ 11/4 . ( - 1 ) ] . 8/33 .

= -11/4 . 8/33 .

= -2/3 .

d, 38/45 - ( 8/45 - 17/51 - 3/11 ) .

= 38/45 - 8/45 + 17/51 + 3/11 .

= 2/3 + 17/51 + 3/11 .

= 374/561 + 187/561 + 153/561 .

= 14/11 .

a) \(\dfrac{15}{12}+\dfrac{5}{13}-\dfrac{3}{12}-\dfrac{18}{13}\)

\(=\left(\dfrac{15}{12}-\dfrac{3}{12}\right)+\left(\dfrac{5}{13}-\dfrac{18}{13}\right)\)

\(=\dfrac{12}{12}+\dfrac{-13}{13}\)

\(=1-1\)

\(=0\)

b) \(\dfrac{5^4\cdot20^4}{25^5\cdot4^5}\)

\(=\dfrac{100^4}{100^5}\)

\(=\dfrac{1}{100}\)

15 tháng 12 2017

a) \(\frac{15}{12}+\frac{5}{13}-\frac{3}{12}-\frac{18}{13}\)

\(=\left(\frac{15}{12}-\frac{3}{12}\right)+\left(\frac{5}{13}-\frac{18}{13}\right)\)

\(=1+\left(-1\right)\)

\(=0\)

b) \(\frac{5^4.20^4}{25^5.4^5}=\frac{\left(20.5\right)^4}{\left(25.4\right)^5}=\frac{100^4}{100^5}=\frac{1}{100}\)

c) \(\frac{8^{10}+4^{10}}{8^4+4^{11}}=\frac{\left(2^3\right)^{10}+\left(2^2\right)^{10}}{\left(2^3\right)^4+\left(2^2\right)^{11}}=\frac{2^{30}+2^{20}}{2^{12}+2^{22}}=\frac{2^{12}.\left(2^{18}+2^8\right)}{2^{12}.\left(1+2^{10}\right)}=\frac{2^{18}+2^8}{1+2^{10}}=256\)

15 tháng 10 2018

\(1,\)

\(a,\dfrac{11}{125}-\dfrac{17}{18}-\dfrac{5}{7}+\dfrac{4}{9}+\dfrac{17}{14}\)

\(=\dfrac{11}{125}+\left(\dfrac{4}{9}-\dfrac{17}{18}\right)+\left(\dfrac{17}{14}-\dfrac{5}{7}\right)\)

\(=\dfrac{11}{125}+\left(\dfrac{-1}{2}\right)+\dfrac{1}{2}\)

\(=\dfrac{11}{125}\)

\(b,-1\dfrac{5}{7}.15+\dfrac{2}{7}.\left(-15\right)+\left(-105\right).\left(\dfrac{2}{3}-\dfrac{4}{5}+\dfrac{1}{7}\right)\)

\(=\dfrac{-12}{7}.15+\dfrac{2}{7}.\left(-15\right)+\left(105\right).\left(\dfrac{2}{3}-\dfrac{4}{5}+\dfrac{1}{7}\right)\)

\(=-15.\left[\dfrac{12}{7}+\dfrac{2}{7}+\left(-5\right).\left(\dfrac{2}{3}-\dfrac{4}{5}+\dfrac{1}{7}\right)\right]\)

\(=-15.\left[2+\left(-5\right).\dfrac{1}{105}\right]\)

\(=-15.\left(2-\dfrac{1}{21}\right)\)

\(=-15.\dfrac{41}{21}=\dfrac{-615}{21}\)

\(2,\)

\(a,\dfrac{11}{13}-\left(\dfrac{5}{42}-x\right)=-\left(\dfrac{15}{28}-\dfrac{11}{13}\right)\)

\(\Leftrightarrow\dfrac{11}{13}-\dfrac{5}{42}+x=\dfrac{-15}{28}+\dfrac{11}{13}\)

\(\Leftrightarrow x=\dfrac{-15}{28}+\dfrac{11}{13}-\dfrac{11}{13}+\dfrac{5}{42}\)

\(\Leftrightarrow x=\left(\dfrac{11}{13}-\dfrac{11}{13}\right)+\left(\dfrac{5}{42}+\dfrac{-15}{28}\right)\)

\(\Leftrightarrow x=\dfrac{5}{12}\)

Vậy \(x=\dfrac{5}{12}\)

\(b,\left|x+\dfrac{4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\)

\(\Leftrightarrow\left|x+\dfrac{4}{15}\right|-3,75=-2,15\)

\(\Leftrightarrow\left|x+\dfrac{4}{15}\right|=-2,15+3,75=1,6=\dfrac{16}{10}=\dfrac{8}{5}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{4}{15}=\dfrac{8}{5}\\x+\dfrac{4}{15}=\dfrac{-8}{5}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{5}-\dfrac{4}{15}=\dfrac{4}{3}\\x=\dfrac{-8}{5}-\dfrac{4}{15}=\dfrac{-28}{15}\end{matrix}\right.\)

Vậy \(x\in\left\{\dfrac{4}{3};\dfrac{-28}{15}\right\}\)

\(c,7^{x+2}+2.7^{x-1}=345\)

\(\Leftrightarrow7^{x-1}.\left(7^3+2\right)=345\)

\(\Leftrightarrow7^{x-1}.\left(343+2\right)=345\)

\(\Leftrightarrow7^{x-1}.345=345\)

\(\Leftrightarrow7^{x-1}=345:345=1\)

\(\Leftrightarrow x-1=0\)

\(x=0+1=1\)

Vậy \(x=1\)

19 tháng 6 2018

8)\(\frac{4}{9}:\left(-\frac{1}{7}\right)+6\frac{5}{9}:\left(-\frac{1}{7}\right)\)

=\(\frac{4}{9}:\left(-\frac{1}{7}\right)+\frac{59}{9}:\left(-\frac{1}{7}\right)\)

=\(\left(\frac{4}{9}+\frac{59}{9}\right).\left(-7\right)\)

=7.(-7)

=-49

Số thập phân hữu hạn là mấy số thập phân không có dấu .... ở đuôi ý bạn ạ.

Còn bài này mình không hiểu rõ đề bài mấy bạn ạ

28/25; 47/20; 15/2; -9/4; -7/4; 9/8

9 tháng 11 2018

a.\(\dfrac{17}{15}\div\dfrac{4}{3}=\dfrac{17}{20}\)

b.\(\dfrac{-12}{21}\div\dfrac{34}{43}=\dfrac{-86}{119}\)

c.\(\dfrac{-5}{9}\times\dfrac{3}{11}+\dfrac{13}{18}\times\dfrac{3}{11}\)

=\(\dfrac{3}{11}\times(\dfrac{-5}{9}+\dfrac{13}{18})=\dfrac{3}{11}\times\dfrac{1}{6}=\dfrac{1}{22}\)

d.\(\dfrac{-2}{9}\times\dfrac{5}{11}+\dfrac{-16}{9}\times\dfrac{5}{11}=\dfrac{5}{11}\times(\dfrac{-2}{9}+\dfrac{-16}{9})\)

=\(\dfrac{5}{11}\times(-2)=\dfrac{-10}{11}\)