\(\frac{11}{12}-\left(\frac{2}{5}+x\right)=\frac{2}{3}\cdot\left(6x+1\right)\)
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6 tháng 4 2018

\(\frac{11}{12}-\left(\frac{2}{5}+x\right)=\frac{2}{3}.\left(6x+1\right)\)

\(\Rightarrow\frac{11}{12}-\frac{2}{5}-x=\frac{2}{3}.6x+\frac{2}{3}\)

\(\Rightarrow\frac{55-24}{60}-x=4x+\frac{2}{3}\)

\(\Rightarrow\frac{31}{60}-x=4x+\frac{2}{3}\)

\(\Rightarrow\frac{31}{60}-\frac{2}{3}=4x+x=5x\)

\(\Rightarrow5x=-\frac{11}{60}\)

\(\Rightarrow x=\frac{-11}{300}\)

1 tháng 4 2020

vô tù ko e

13 tháng 7 2019

#)Giải :

a)\(2009^{\left(1000-1^3\right)\left(1000-2^3\right)...\left(1000-15^3\right)}=2009^{\left(1000-1^3\right)...\left(1000-10^3\right)...\left(1000-15^3\right)}=2009^0=1\)

b)\(\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)=\left(\frac{1}{125}-\frac{1}{1^3}\right)...\left(\frac{1}{125}-\frac{1}{5^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)=\left(\frac{1}{125}-\frac{1}{1^3}\right)...0...\left(\frac{1}{125}-\frac{1}{25^3}\right)=0\)

\(a,\left[\left(-\frac{1}{2}\right)^3-\left(\frac{3}{4}\right)^3.\left(-2\right)^2\right]:\left[2.\left(-1\right)^5+\left(\frac{3}{4}\right)^2-\frac{3}{8}\right]\)

\(=\left[\left(-\frac{1}{8}\right)-\frac{27}{64}.4\right]:\left[2.\left(-1\right)+\frac{9}{16}-\frac{3}{8}\right]\)

\(=\left[\left(-\frac{1}{8}-\frac{27}{16}\right)\right]:\left[-2+\frac{9}{16}-\frac{3}{8}\right]\)

\(=\frac{-2-27}{16}:\frac{-32+9-6}{16}\)

\(=-\frac{29}{16}:\frac{-29}{16}=1\)

\(b,\left[\left(\frac{4}{3}\right)^{-2}\left(\frac{3}{2}\right)^4\right]:\left(\frac{3}{2}\right)^6\)

\(=\left(\frac{9}{16}.\frac{81}{16}\right):\frac{729}{64}\)

\(=\frac{729}{64}:\frac{729}{64}=1\)

14 tháng 5 2019

\(\frac{\left(\frac{2}{3}\right)^3\cdot\left(-\frac{3}{4}^2\right)\cdot\left(-1\right)^{2003}}{\left(\frac{2}{5}\right)^2\cdot\left(-\frac{5}{12}\right)^3}\)

\(=\frac{\frac{8}{27}\cdot\frac{9}{16}\cdot\left(-1\right)}{\frac{4}{25}\cdot\left(-\frac{125}{1728}\right)}\)

\(=\frac{-\frac{1}{6}}{-\frac{5}{432}}=-\frac{1}{6}:\left(-\frac{5}{432}\right)=\frac{72}{5}\)

14 tháng 5 2019

\(\left[6.\left(\frac{-1}{3}\right)^2-3.\left(\frac{-1}{3}\right)+1\right]:\left(\frac{-1}{3}-1\right)\)

\(=\left[6.\frac{1}{9}-\left(-1\right)+1\right]:\frac{-4}{3}\)

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

\(=\frac{8}{3}:\frac{-4}{3}=\frac{-24}{12}=-2\)

~ Hok tốt ~

11 tháng 12 2019

a) 12. \(\frac{4}{9}\)+\(\frac{4}{3}\)=\(\frac{16}{3}\)+\(\frac{4}{3}\)=\(\frac{20}{3}\)

b) (\(\frac{-5}{7}\)) . (12,5+1,5)= (\(\frac{-5}{7}\)).14=-10

a) \(12.\left(-\frac{2}{3}\right)^2+\frac{4}{3}=12.\frac{4}{9}+\frac{4}{3}=\frac{16}{3}+\frac{4}{3}=\frac{20}{3}\)

b) \(12,5.\left(-\frac{5}{7}\right)+1,5.\left(-\frac{5}{7}\right)=-\frac{5}{7}.\left(12,5+1,5\right)=-\frac{5}{7}.14=-10\)

c) \(1:\left(\frac{2}{3}-\frac{3}{4}\right)^2=1:\left(-\frac{1}{12}\right)^2=1:\frac{1}{144}=1.144=144\)

d) \(15.\left(-\frac{2}{3}\right)^2-\frac{7}{3}=15.\frac{4}{9}-\frac{7}{3}=\frac{20}{3}-\frac{7}{3}=\frac{13}{3}\)

e) \(\frac{1}{2}\sqrt{64}-\sqrt{\frac{4}{25}}+\left(-1\right)^{2007}=\frac{1}{2}.8-\frac{2}{5}+\left(-1\right)=4-\frac{2}{5}-1=\frac{13}{5}\)

11 tháng 12 2019

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) \(\frac{11}{24}-\frac{5}{41}+\frac{13}{24}+0,5-\frac{36}{41}=\left(\frac{11}{24}+\frac{13}{24}\right)+\left(-\frac{5}{41}-\frac{36}{41}\right)+0,5\)

                                                                    \(=1+\left(-1\right)+0,5\)

                                                                    \(=0,5\)

_Học tốt nha_

11 tháng 12 2019

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

\(\frac{5}{4}\)\(\frac{5}{13}\) - \(\frac{1}{4}\) - \(\frac{18}{13}\)

\(\left(\frac{5}{4}-\frac{1}{4}\right)\)\(\left(\frac{5}{13}-\frac{18}{13}\right)\)

= 1 - 1 = 0

b, \(\frac{11}{24}\)\(\frac{5}{41}\)\(\frac{13}{24}\)+ 0,5 - \(\frac{36}{41}\)

\(\left(\frac{11}{24}+\frac{13}{24}\right)\)\(\left(\frac{5}{41}+\frac{36}{41}\right)\)+ 0,5

= 1 - 1 + 0,5 = 0,5

c,  \(\left(-\frac{3}{4}+\frac{2}{3}\right):\frac{5}{11}+\left(-\frac{1}{4}+\frac{1}{3}\right):\frac{5}{11}\)

=\(\left(-\frac{3}{4}+\frac{2}{3}\right).\frac{11}{5}+\left(-\frac{1}{4}+\frac{1}{3}\right).\frac{5}{11}\)

\(\frac{11}{5}.\left(-\frac{3}{4}+\frac{2}{3}-\frac{1}{4}+\frac{1}{3}\right)\)

\(\frac{11}{5}.\left[\left(-\frac{3}{4}-\frac{1}{4}\right)+\left(\frac{2}{3}+\frac{1}{3}\right)\right]\)

=  \(\frac{11}{5}.\left[\left(-1\right)+1\right]\)

= 0

d, \(\left(-3\right)^2.\left(\frac{3}{4}-0,25\right)-\left(3\frac{1}{2}-1\frac{1}{2}\right)\)

\(9.\left(0,75-0,25\right)-2\)

= 9. 0,5 - 2 = 2,5

e, \(\frac{13}{25}+\frac{6}{41}-\frac{38}{25}+\frac{35}{41}-\frac{1}{2}\)

\(\left(\frac{13}{25}-\frac{38}{25}\right)+\left(\frac{6}{41}+\frac{35}{41}\right)-\frac{1}{2}\)

= -1 + 1 - \(\frac{1}{2}\)

\(-\frac{1}{2}\)