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28 tháng 4 2017

Bài 1:

a) \(\left(\dfrac{3}{8}+\dfrac{-3}{4}+\dfrac{7}{12}\right):\dfrac{5}{6}+\dfrac{1}{2}\)

\(=\left(\dfrac{9}{24}+\dfrac{-18}{24}+\dfrac{14}{24}\right):\dfrac{5}{6}+\dfrac{1}{2}\)

\(=\dfrac{5}{24}:\dfrac{5}{6}+\dfrac{1}{2}\)

\(=\dfrac{5}{24}.\dfrac{6}{5}+\dfrac{1}{2}\)

\(=\dfrac{1}{4}+\dfrac{1}{2}\)

\(=\dfrac{1}{4}+\dfrac{2}{4}\)

\(=\dfrac{3}{4}\)

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

\(=\dfrac{1}{2}+\dfrac{3}{4}-\dfrac{3}{4}+\dfrac{4}{5}\)

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

\(=\dfrac{1}{2}+\dfrac{4}{5}\)

\(=\dfrac{5}{10}+\dfrac{8}{10}\)

\(=\dfrac{9}{5}\)

c) \(6\dfrac{5}{12}:2\dfrac{3}{4}+11\dfrac{1}{4}.\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)

\(=\dfrac{77}{12}:\dfrac{11}{4}+\dfrac{42}{4}.\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)

\(=\dfrac{77}{12}.\dfrac{4}{11}+\dfrac{42}{4}.\left(\dfrac{5}{15}+\dfrac{3}{15}\right)\)

\(=\dfrac{7}{3}+\dfrac{42}{4}.\dfrac{8}{15}\)

\(=\dfrac{7}{3}+\dfrac{14.2}{1.3}\)

\(=\dfrac{7}{3}+\dfrac{28}{3}\)

\(=\dfrac{35}{3}\)

d) \(\left(\dfrac{7}{8}-\dfrac{3}{4}\right).1\dfrac{1}{3}-\dfrac{2}{7}.\left(3,5\right)^2\)

\(=\left(\dfrac{7}{8}-\dfrac{6}{8}\right).\dfrac{4}{3}-\dfrac{2}{7}.12\dfrac{1}{4}\)

\(=\dfrac{1}{8}.\dfrac{4}{3}-\dfrac{2}{7}.\dfrac{49}{4}\)

\(=\dfrac{1}{6}-\dfrac{7}{2}\)

\(=\dfrac{1}{6}-\dfrac{21}{6}\)

\(=\dfrac{-10}{3}\)

e) \(\left(\dfrac{3}{5}+0,415-\dfrac{3}{200}\right).2\dfrac{2}{3}.0,25\)

\(=\left(\dfrac{3}{5}+\dfrac{83}{200}-\dfrac{3}{200}\right).\dfrac{8}{3}.\dfrac{1}{4}\)

\(=\left(\dfrac{120}{200}+\dfrac{83}{200}-\dfrac{3}{200}\right).\dfrac{8}{3}.\dfrac{1}{4}\)

\(=1.\dfrac{8}{3}.\dfrac{1}{4}\)

\(=\dfrac{2}{3}\)

f) \(\dfrac{5}{16}:0,125-\left(2\dfrac{1}{4}-0,6\right).\dfrac{10}{11}\)

\(=\dfrac{5}{16}:\dfrac{1}{8}-\left(\dfrac{9}{4}-\dfrac{3}{5}\right).\dfrac{10}{11}\)

\(=\dfrac{5}{16}.\dfrac{8}{1}-\left(\dfrac{45}{20}-\dfrac{12}{20}\right).\dfrac{10}{11}\)

\(=\dfrac{5}{2}-\dfrac{33}{20}.\dfrac{10}{11}\)

\(=\dfrac{5}{2}-\dfrac{3}{2}\)

\(=\dfrac{2}{2}=1\)

g) \(0,25:\left(10,3-9,8\right)-\dfrac{3}{4}\)

\(=\dfrac{1}{4}:\dfrac{1}{2}-\dfrac{3}{4}\)

\(=\dfrac{1}{4}.\dfrac{2}{1}-\dfrac{3}{4}\)

\(=\dfrac{1}{2}-\dfrac{3}{4}\)

\(=\dfrac{2}{4}-\dfrac{3}{4}\)

\(=\dfrac{-1}{4}\)

h) \(1\dfrac{13}{15}.0,75-\left(\dfrac{11}{20}+20\%\right):\dfrac{7}{3}\)

\(=\dfrac{28}{15}.\dfrac{3}{4}-\left(\dfrac{11}{20}+\dfrac{1}{5}\right):\dfrac{7}{3}\)

\(=\dfrac{7}{5}-\left(\dfrac{11}{20}+\dfrac{4}{20}\right):\dfrac{7}{3}\)

\(=\dfrac{7}{5}-\dfrac{3}{4}:\dfrac{7}{3}\)

\(=\dfrac{7}{5}-\dfrac{9}{28}\)

\(=\dfrac{196}{140}-\dfrac{45}{140}\)

\(=\dfrac{151}{140}\)

i) \(\dfrac{\left(\dfrac{1}{2-0,75}\right).\left(0,2-\dfrac{2}{5}\right)}{\dfrac{5}{9}-1\dfrac{1}{12}}\)

\(=\dfrac{\left(\dfrac{1}{1,25}\right).\left(\dfrac{1}{5}-\dfrac{2}{5}\right)}{\dfrac{5}{9}-\dfrac{13}{12}}\)

\(=\dfrac{\dfrac{1}{1,25}.\dfrac{-1}{5}}{\dfrac{20}{36}-\dfrac{39}{36}}\)

\(=\dfrac{\dfrac{-1}{6,25}}{\dfrac{-19}{36}}\)

k) \(\dfrac{\dfrac{2}{3}+\dfrac{2}{7}-\dfrac{1}{14}}{-1-\dfrac{3}{7}+\dfrac{3}{28}}\)

\(=\dfrac{\dfrac{2}{3}+\dfrac{2}{7}-\dfrac{2}{28}}{-\dfrac{3}{3}-\dfrac{3}{7}+\dfrac{3}{28}}\)

\(=\dfrac{2\left(\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{28}\right)}{\left(-3\right)\left(\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{28}\right)}\)

\(=-\dfrac{2}{3}\)

29 tháng 4 2017

\(A=0,7.2\dfrac{2}{3}.20.0,375.\dfrac{5}{28}\)

\(A=\dfrac{7}{10}.\dfrac{8}{3}.20.\dfrac{3}{8}.\dfrac{5}{28}\)

\(A=\left(\dfrac{7}{10}.\dfrac{5}{28}\right).\left(\dfrac{8}{3}.\dfrac{3}{8}\right).20\)

\(A=\dfrac{1}{8}.1.20\)

\(A=\dfrac{20}{8}=\dfrac{5}{2}\)

\(B=\left(9\dfrac{30303}{80808}+7\dfrac{303030}{484848}\right)+4,03\)

\(B=\left(9\dfrac{3}{8}+7\dfrac{5}{8}\right)+4,03\)

\(B=\left[\left(9+7\right)+\left(\dfrac{3}{8}+\dfrac{5}{8}\right)\right]+4,03\)

\(B=\left(16+1\right)+4,03\)

\(B=17+4,03\)

\(B=21,03\)

\(C=\left(9,75.21\dfrac{3}{7}+\dfrac{39}{4}.18\dfrac{4}{7}\right).\dfrac{15}{78}\)

\(C=\left(\dfrac{39}{4}.\dfrac{150}{7}+\dfrac{39}{4}.\dfrac{130}{7}\right).\dfrac{15}{78}\)

\(C=\dfrac{39}{4}.\left(\dfrac{150}{7}+\dfrac{130}{7}\right).\dfrac{15}{78}\)

\(C=\dfrac{39}{4}.40.\dfrac{15}{78}\)

\(C=390.\dfrac{15}{78}\)

\(C=75\)

a: =11+3/4-6-5/6+4+1/2+1+2/3

=10+9/12-10/12+6/12+8/12

=10+13/12=133/12

b: \(=2+\dfrac{17}{20}-1-\dfrac{11}{15}+2+\dfrac{3}{20}\)

=3-11/15

=34/15

c: \(=\dfrac{31}{7}:\left(\dfrac{7}{5}\cdot\dfrac{31}{7}\right)\)

\(=\dfrac{31}{7}:\dfrac{31}{5}=\dfrac{5}{7}\)

d: \(=\dfrac{29}{8}\cdot\dfrac{36}{29}\cdot\dfrac{15}{23}\cdot\dfrac{23}{5}=\dfrac{9}{2}\cdot3=\dfrac{27}{2}\)

2:

a: =>2/3:x=1,4-2,4=-1

=>x=-2/3

b: =>x/5=25/30-19/30=6/30=1/5

=>x=1

3:

Số học sinh giỏi là 40*1/4=10 bạn

Số học sinh khá là 30*3/5=18 bạn

Số học sinh TB là 30-18=12 bạn

 

11 tháng 4 2022

a, = (58/9 + 7/11) - (40/9 - 26/11)

= 701/99 - 206/99

= 5

b, = 51/5 - 11/2 . 60/11 + 3 : 3/20

= 51/5 - 30 + 20

= -99/5 + 20

= 1/5

c, = 19/4 + (-0,37) + 1/8 + (-1,8) + (-2,5) + 37/12

= 219/50 + -67/40 + 7/12

= 1973/600

14 tháng 9 2021

\(a.\dfrac{-42}{12}+\dfrac{9}{12}-\dfrac{17}{12}=-\dfrac{25}{6}\)

\(b.-\dfrac{1}{12}-\dfrac{5}{4}+\dfrac{1}{3}=-\dfrac{1}{12}-\dfrac{15}{12}+\dfrac{4}{12}=-1\)

14 tháng 9 2021

a) \(\dfrac{-7}{2}+\dfrac{3}{4}-\dfrac{17}{12}=\dfrac{-42}{12}+\dfrac{9}{12}-\dfrac{17}{12}=\dfrac{-42+9-17}{12}=-\dfrac{50}{12}=-\dfrac{25}{6}\)

b)\(-\dfrac{1}{12}-\left(2\dfrac{5}{8}-\dfrac{1}{3}\right)=-\dfrac{1}{12}-\left(\dfrac{21}{8}-\dfrac{1}{3}\right)=-\dfrac{2}{24}-\left(\dfrac{63}{24}-\dfrac{8}{24}\right)\)

\(=-\dfrac{2}{24}-\dfrac{63}{24}+\dfrac{8}{24}=-\dfrac{57}{24}=-\dfrac{19}{8}\)

24 tháng 7 2021

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24 tháng 7 2021

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15 tháng 12 2021

\(a.=\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{5}{3}+\dfrac{3}{2}+\dfrac{7}{3}-\dfrac{5}{2}=\dfrac{1+3-5}{2}-\dfrac{2+5-7}{3}=\dfrac{-1}{2}\)

\(b.\left(\dfrac{3}{4}-1\dfrac{1}{6}\right)^2:\sqrt{\dfrac{25}{144}}=\left(-\dfrac{5}{12}\right)^2:\dfrac{5}{12}=\dfrac{5}{12}\)

15 tháng 12 2021

thanks nhìu

11 tháng 2 2018

\(B=\frac{1-\frac{1}{\sqrt{49}}+\frac{1}{49}-\frac{1}{\left(7\sqrt{7}\right)^2}}{\frac{\sqrt{64}}{2}-\frac{4}{7}+\frac{2^2}{7^2}-\frac{4}{343}}\)

\(B=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{\frac{8}{2}-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}\)

\(B=\frac{\frac{343}{343}-\frac{49}{343}+\frac{7}{343}-\frac{1}{343}}{4-\frac{4}{7}+\frac{28}{343}-\frac{4}{343}}\)

\(B=\frac{\frac{300}{343}}{\frac{28}{7}-\frac{4}{7}+\frac{24}{343}}\)

\(B=\frac{\frac{300}{343}}{\frac{24}{7}+\frac{24}{343}}\)

\(B=\frac{\frac{300}{343}}{\frac{1323}{343}+\frac{24}{343}}\)

\(B=\frac{300}{343}:\frac{1347}{343}\)

\(B=\frac{100}{449}\)

11 tháng 2 2018

\(A=\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\frac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)

\(A=\frac{2^{12}.3^5-2^{12}.3^6}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}.7^6}{5^9.7^3+5^9.2^3.7^3}\)

\(A=\frac{2^{12}.3^5\left(1-3\right)}{2^{12}.3^5.\left(3+1\right)}-\frac{5^{10}.7^3.\left(1-7^3\right)}{5^9.7^3.\left(1+8\right)}\)

\(A=\frac{-2}{4}-\frac{5.\left(-342\right)}{9}\)

\(A=\frac{-1}{2}+\frac{1710}{9}\)

\(A=\frac{-1}{2}+190\)

\(A=\frac{-1}{2}+\frac{380}{2}\)

\(A=\frac{379}{2}\)

28 tháng 5 2022

`1//([-1]/2)^2 . |+8|-(-1/2)^3:|-1/16|=1/4 .8+1/8 .16=2+2=4`

`2//|-0,25|-(-3/2)^2:1/4+3/4 .2017^0=0,25-2,25.4+0,75.1=0,25-9+0,75=-8,75+0,75-8`

`3//|2/3-5/6|.(3,6:2 2/5)^3=|-1/6|.(3/2)^3=1/6 . 27/8=9/16`

`4//|(-0,5)^2+7/2|.10-(29/30-7/15):(-2017/2018)^0=|1/4+7/2|.10-1/2:1=|15/4|.10-1/2=15/4 .10-1/2=75/2-1/2=37`

`5// 8/3+(3-1/2)^2-|[-7]/3|=8/3+(5/2)^2-7/3=8/3+25/4-7/3=107/12-7/3=79/12`

a: \(A=\dfrac{7}{12}+\dfrac{5}{72}-\dfrac{11}{36}=\dfrac{42}{72}+\dfrac{5}{72}-\dfrac{22}{72}=\dfrac{25}{72}\)

b: \(B=\dfrac{8+5}{10}:\dfrac{-5}{13}=\dfrac{13}{10}\cdot\dfrac{13}{-5}=-\dfrac{169}{100}\)

c: \(C=\left(\dfrac{88}{132}-\dfrac{33}{132}+\dfrac{60}{132}\right):\left(\dfrac{55}{132}+\dfrac{132}{132}-\dfrac{84}{132}\right)\)

\(=\dfrac{88-33+60}{55+132-84}=\dfrac{115}{103}\)