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1 tháng 3

1+1

 

a: \(\dfrac{4}{7}\cdot\dfrac{-28}{3}=\dfrac{4\cdot\left(-28\right)}{7\cdot3}=\dfrac{-112}{21}=\dfrac{-16}{3}\)

b: \(\left(-15\right)\cdot\dfrac{3}{5}=\dfrac{-15\cdot3}{5}=\dfrac{-45}{5}=-9\)

c: \(\dfrac{9}{5}\cdot\dfrac{-20}{7}=\dfrac{-20}{5}\cdot\dfrac{9}{7}=-4\cdot\dfrac{9}{7}=-\dfrac{36}{7}\)

d: \(\dfrac{-10}{9}\cdot\dfrac{-3}{20}=\dfrac{3}{9}\cdot\dfrac{10}{20}=\dfrac{1}{2}\cdot\dfrac{1}{3}=\dfrac{1}{6}\)

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

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)\(\dfrac{-10}{11}.\dfrac{8}{9}+\dfrac{7}{18}.\dfrac{10}{11}\)

=\(\dfrac{10}{11}.\dfrac{-8}{9}+\dfrac{7}{18}.\dfrac{10}{11}\)

=\(\dfrac{10}{11}(\dfrac{-8}{9}+\dfrac{7}{18})\)

=\(\dfrac{10}{11}.\dfrac{-1}{2}\)

=\(\dfrac{-5}{11}\)

7 tháng 8

b; 

B = \(\dfrac{3}{14}\) : \(\dfrac{1}{28}\) - \(\dfrac{13}{21}\)\(\dfrac{1}{28}\) + \(\dfrac{29}{42}\) : \(\dfrac{1}{28}\) - 8

B = (\(\dfrac{3}{14}\) - \(\dfrac{13}{21}\) + \(\dfrac{29}{42}\)) - 8

B = (\(\dfrac{9}{42}\) - \(\dfrac{26}{42}\) + \(\dfrac{29}{42}\)) - 8

B = (\(\dfrac{-17}{42}\) + \(\dfrac{29}{42}\)) - 8

B = \(\dfrac{2}{7}\) - 8

B = \(\dfrac{2}{7}-\dfrac{56}{7}\)

B = - \(\dfrac{54}{7}\)

24 tháng 1 2022

=\(\left[\dfrac{\left(0,4.2\right)^5}{\left(0,4\right)^6}+\dfrac{2^9.2^6.3^8}{\left(3.2\right)^6.2^9}\right]=\left[\dfrac{\left(0,4\right)^5.2^5}{\left(0,4\right)^6}+\dfrac{2^6.3^8}{3^6.2^6}\right]\)

=\(\left[\dfrac{2^5}{0,4}+3^2\right]\)

=\(\left[80+9\right]=89\)

24 tháng 1 2022

\(\left[\dfrac{\left(2.0,4\right)^5}{0,4,0,4^5}+\dfrac{2^{15}.3^8}{3^6.2^6.2^9}\right]\div\dfrac{3^{20}.5^{30}}{3^{15}.5^{30}}\)

\(=\left[\dfrac{2^5.0.4^5}{0,4.0,4^5}+\dfrac{2^{15}.3^8}{3^6.2^{15}}\right]\div3^5\)

\(=\left[\dfrac{2^5}{0,4}+3^2\right]\div243\)

\(=80+\left(3^5\div3^2\right)\)

\(=80+3^3\)

\(=80+27\)

\(=107\)

13 tháng 9 2017

a/ \(\dfrac{\left(1+2+.....+100\right)\left(\dfrac{1}{3}-\dfrac{1}{5}-\dfrac{1}{7}-\dfrac{1}{9}\right)\left(6,3.12-21.36\right)}{\dfrac{1}{2}+\dfrac{1}{3}+.......+\dfrac{1}{100}}\)

\(=\dfrac{\left(1+2+3+.....+100\right)\left(\dfrac{1}{3}-\dfrac{1}{5}-\dfrac{1}{7}-\dfrac{1}{9}\right).0}{\dfrac{1}{2}+\dfrac{1}{3}+.......+\dfrac{1}{100}}\)

\(=\dfrac{0}{\dfrac{1}{2}+\dfrac{1}{3}+.....+\dfrac{1}{100}}\)

\(=0\)

13 tháng 9 2017

bn có chép sai đề bài ko vậy

20 tháng 12 2021

e: \(=\dfrac{5^{30}\cdot3^{20}}{3^{15}\cdot5^{30}}=3^5=243\)

10 tháng 7 2017

Cái này bn lầy máy tính ra tính tí là xong thôi

5 tháng 5 2018

a) \(2\dfrac{3}{4}.\left(-0,4\right)-1\dfrac{3}{5}.2,75+\left(-1,2\right):\dfrac{4}{11}\)

= \(2,75.\left(-0,4\right)-\left(1,6\right).\left(2,75\right)+\left(-1,2\right).\dfrac{11}{4}\)

= \(2,75.\left(-0,4\right)-\left(1,6\right).\left(2,75\right)+\left(-1,2\right).\left(2,75\right)\)

= \(2,75.\left\{\left(-0,4\right)-\left(1,6\right)+\left(-1,2\right)\right\}\)

= \(2,75.\left(-3,2\right)\)

= \(-8,8\)

b) \(1,4.\dfrac{15}{49}-\left(\dfrac{4}{5}+\dfrac{2}{3}\right):2\dfrac{1}{5}\)

= \(\dfrac{7}{5}.\dfrac{15}{49}-\left(\dfrac{4}{5}+\dfrac{2}{3}\right):\dfrac{11}{5}\)

= \(\dfrac{7}{5}.\dfrac{15}{49}-\dfrac{22}{15}.\dfrac{5}{11}\)

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

= \(-\dfrac{5}{21}\)

c) \(\left(-3,2\right).\dfrac{15}{64}+\left(0,8-2\dfrac{4}{15}\right):3\dfrac{2}{3}\)

= \(-\dfrac{16}{5}.\dfrac{15}{64}+\left(\dfrac{4}{5}-2\dfrac{4}{15}\right):\dfrac{11}{3}\)

= \(-\dfrac{16}{5}.\dfrac{15}{64}+\left(-\dfrac{22}{15}\right).\dfrac{3}{11}\)

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

= \(-\dfrac{23}{20}\)

d) \(0,02.\dfrac{-25}{2}+\dfrac{3}{8}+\left(-2\dfrac{9}{20}\right).\dfrac{2}{7}\)

= \(\dfrac{1}{50}.\dfrac{-25}{2}+\dfrac{3}{8}+\left(-\dfrac{49}{20}\right).\dfrac{2}{7}\)

=\(\left(-\dfrac{1}{4}\right)+\dfrac{3}{8}+\left(-\dfrac{7}{10}\right)\)

= \(\dfrac{1}{8}+\left(-\dfrac{7}{10}=\right)\)

= \(-\dfrac{23}{40}\)

e) \(34\%:\dfrac{51}{16}-3\dfrac{7}{9}.6,5-\left(0,4\right)^2\)

= \(\dfrac{17}{50}.\dfrac{16}{51}-\dfrac{34}{9}.\dfrac{13}{2}-\dfrac{4}{25}\)

= \(\dfrac{8}{75}-\dfrac{221}{9}-\dfrac{4}{15}\)

= \(-\dfrac{5501}{225}\)

5 tháng 8 2021

a, \(\dfrac{5}{7}+\left(\dfrac{3}{5}+\dfrac{-5}{7}\right)\)
\(=\dfrac{5}{7}+\dfrac{-5}{7}+\dfrac{3}{5} =0+\dfrac{3}{5}=\dfrac{3}{5}\)

b, \(=\dfrac{-3}{4}-\dfrac{15}{14}:\dfrac{-5}{7}+\left(-1\right)^2=\dfrac{-3}{4}-\dfrac{-3}{2}+1=\dfrac{-3}{4}-\dfrac{-6}{4}+1=\dfrac{3}{4}+1=\dfrac{7}{4}\)

 c, \(\dfrac{-5}{9}+\left(\dfrac{-2}{3}\right)^2.\left(20\%-1.2\right)=\dfrac{-5}{9}+\dfrac{4}{9}x\left(\dfrac{1}{5}-\dfrac{6}{5}\right)=\dfrac{-5}{9}+\dfrac{4}{9}x\left(-1\right)=\dfrac{-5}{9}+\dfrac{-4}{9}=-1\)

5 tháng 8 2021

Bài 1:

a) \(\dfrac{5}{7}+\left(\dfrac{3}{5}+\dfrac{-5}{7}\right)\)\(=\left(\dfrac{5}{7}+\dfrac{-5}{7}\right)+\dfrac{3}{5}\)\(=0+\dfrac{3}{5}=\dfrac{3}{5}\)

b) \(\dfrac{-3}{4}-\dfrac{15}{14}:\dfrac{-5}{7}+\left(-1\right)^2\)\(=\dfrac{-3}{4}-\dfrac{15}{14}:\dfrac{-5}{7}+1\)\(=\dfrac{-3}{4}-\dfrac{-3}{2}+1\)

                                            \(=\dfrac{3}{4}+1\)\(=\dfrac{7}{4}\)

a: \(A=\dfrac{9^4}{3^2}=\dfrac{\left(3^2\right)^4}{3^2}=\dfrac{3^8}{3^2}=3^6\)=729

b: \(B=81\left(\dfrac{5}{3}\right)^4=81\cdot\dfrac{5^4}{3^4}=\dfrac{81}{3^4}\cdot5^4=5^4=625\)

c: \(C=\left(\dfrac{4}{7}\right)^{-4}\cdot\left(\dfrac{2}{7}\right)^3\)

\(=\left(\dfrac{7}{4}\right)^4\cdot\left(\dfrac{2}{7}\right)^3\)

\(=\dfrac{7^4}{4^4}\cdot\dfrac{2^3}{7^3}\)

\(=\dfrac{2^3}{4^4}\cdot7\)

\(=\dfrac{2^3}{2^8}\cdot7=\dfrac{7}{2^5}=\dfrac{7}{32}\)

d: \(D=7^{-6}\cdot\left(\dfrac{2}{3}\right)^0\left(\dfrac{7}{5}\right)^6\)

\(=7^{-6}\left(\dfrac{7}{5}\right)^6\)

\(=\dfrac{1}{7^6}\cdot\dfrac{7^6}{5^6}=\dfrac{1}{5^6}=\dfrac{1}{15625}\)

e: \(E=8^3:\left(\dfrac{2}{3}\right)^5\cdot\left(\dfrac{1}{3}\right)^2\)

\(=2^6:\dfrac{2^5}{3^5}\cdot\dfrac{1}{3^2}\)

\(=2^6\cdot\dfrac{3^5}{2^5}\cdot\dfrac{1}{3^2}\)

\(=\dfrac{2^6}{2^5}\cdot\dfrac{3^5}{3^2}=3^3\cdot2=54\)

f: \(F=\left(\dfrac{7}{9}\right)^{-2}\cdot\left(\dfrac{1}{\sqrt{3}}\right)^8\)

\(=\left(\dfrac{9}{7}\right)^2\cdot\left(\dfrac{1}{3}\right)^4\)

\(=\dfrac{9^2}{7^2}\cdot\dfrac{1}{3^4}=\dfrac{9^2}{3^4}\cdot\dfrac{1}{7^2}=\dfrac{81}{81}\cdot\dfrac{1}{49}=\dfrac{1}{49}\)

g: \(G=\left(-\dfrac{4}{5}\right)^{-2}\cdot\left(\dfrac{2}{5}\right)^2\cdot\left(\sqrt{2}\right)^3\)

\(=\left(-\dfrac{5}{4}\right)^2\cdot\left(\dfrac{2}{5}\right)^2\cdot2\sqrt{2}\)

\(=\dfrac{25}{16}\cdot\dfrac{4}{25}\cdot2\sqrt{2}=\dfrac{4}{16}\cdot2\sqrt{2}=\dfrac{8\sqrt{2}}{16}=\dfrac{\sqrt{2}}{2}\)

10 tháng 10 2018

a, \(125^3:5^7=\left(5^3\right)^3:5^7=5^9:5^7=5^2\)

b, \(\left(\dfrac{2}{7}\right)^{18}:\left(\dfrac{4}{49}\right)^5:\left(\dfrac{8}{343}\right)^2\)

= \(\left(\dfrac{2}{7}\right)^{18}:\left(\dfrac{2^2}{7^2}\right)^5:\left(\dfrac{2^3}{7^3}\right)^2\)

= \(\left(\dfrac{2}{7}\right)^{18}:\left[\left(\dfrac{2}{7}\right)^2\right]^5:\left[\left(\dfrac{2}{7}\right)^3\right]^2\)

=\(\left(\dfrac{2}{7}\right)^{18}:\left(\dfrac{2}{7}\right)^{10}:\left(\dfrac{2}{7}\right)^6\)

= \(\left(\dfrac{2}{7}\right)^{18-10-6}=\left(\dfrac{2}{7}\right)^2\)

c, \(3-\left(\dfrac{-7}{9}\right)^0+\left(\dfrac{1}{3}\right)^5.3^5\)

= 3 - 1 +\(\left[\left(\dfrac{1}{3}\right)^5.3^5\right]\)

= 2 + 1=3

d, \(\dfrac{45^{10}.5^{20}}{75^{15}}=\dfrac{\left(9.5\right)^{10}.5^{20}}{\left(25.3\right)^{15}}=\dfrac{\left(3^2\right)^{10}.5^{10}.5^{20}}{\left(5^2\right)^{15}.3^{15}}\)

= \(\dfrac{3^{20}.5^{30}}{5^{30}.3^{15}}=3^5\)