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5 tháng 11 2019

    (1/3)50. (-9)25 - 2/3 : 4

= (1/3)50 . [(-3)2]25 - 2/3 . 1/4

=  (1/3)50.(-3)50- 1/6

= (1/3 . -3 )50 - 1/6

= (-1)50- 1/6

= 1 - 1/6

= 5/6

10 tháng 11 2019

theo đề ta có

=\(\left(\frac{1}{3^{ }}\right)^{50}.\left(-9\right)^{25}-\frac{2}{3}.\frac{1}{4}\)

=\(\left(\frac{1}{3}\right)^{50}.\left(\frac{1}{3}\right)^{25}.\left(-27\right)^{25}-\frac{1}{6}\)

=\(\left(\frac{1}{3}\right)^{50+25}.\)

15 tháng 3 2020

Ta có : \(\left(2^2:\frac{4}{3}-\frac{1}{2}\right).\frac{6}{5}-17\)

=\(=\left(4.\frac{3}{4}-\frac{1}{2}\right).\frac{6}{5}-17\)

\(=\frac{5}{2}.\frac{6}{5}-17\)

\(=3-17=-14\)

Tụi quá mới lớp 5 thui

5 tháng 12 2015

a) (1/3)50 . (-9)25 .2/3:4

= (1/3)50 . [-(3)50] . 2/3 . 1/4

= - (1/3.3)50 . 1/6

= -1 . 1/6 = -1/6

b) (24.26)/(25)2 - (25.153)/(63.102)

= 210/210 - (25.33.53)/(23.33.22.52)

= 1- (25.33.53)/(25.33.52)

= 1- 5

= -4

a: \(=\dfrac{5}{3}\left(-16-\dfrac{2}{7}+28+\dfrac{2}{7}\right)=\dfrac{5}{3}\cdot12=20\)

b: \(=\left(4\cdot\dfrac{3}{4}-\dfrac{1}{2}\right)\cdot\dfrac{6}{5}-17=\dfrac{1}{2}\cdot\dfrac{6}{5}-17=\dfrac{3}{5}-17=-\dfrac{82}{5}\)

c: \(=-\left(\dfrac{1}{3}\right)^{50}\cdot3^{50}-\dfrac{2}{3}\cdot\dfrac{1}{4}=-1-\dfrac{1}{6}=-\dfrac{7}{6}\)

e: \(=5.7\left(-6.5-3.5\right)=-5.7\cdot10=-57\)

15 tháng 10 2016

\(E=\frac{0,8:\left(\frac{4}{5}.1,25\right)}{0,64-\frac{1}{25}}+\frac{\left(1,08-\frac{2}{25}\right):\frac{4}{7}}{\left(6\frac{5}{9}-3\frac{1}{4}\right).2\frac{2}{17}}+\left(1,2.0,5\right):\frac{4}{5}\)

\(E=\frac{\frac{4}{5}:\frac{4}{5}:1,25}{\frac{16}{25}-\frac{1}{25}}+\frac{\left(\frac{27}{25}-\frac{2}{25}\right).\frac{7}{4}}{\left(\frac{59}{9}-\frac{13}{4}\right).\frac{36}{17}}+\frac{6}{5}.\frac{1}{2}.\frac{5}{4}\)

\(E=\frac{1:\frac{5}{4}}{\frac{3}{5}}+\frac{1.\frac{7}{4}}{\frac{119}{36}.\frac{36}{17}}+\frac{3}{4}\)

\(E=\frac{4}{5}.\frac{5}{3}+\frac{\frac{7}{4}}{7}+\frac{3}{4}\)

\(E=\frac{4}{3}+\frac{7}{4}.\frac{1}{7}+\frac{3}{4}\)

\(E=\frac{4}{3}+\frac{1}{4}+\frac{3}{4}\)

\(E=\frac{4}{3}+1=\frac{7}{3}\)

14 tháng 10 2016

Bấm máy tính:

E = \(\frac{4}{3}+\frac{1}{4}+\frac{3}{5}:\frac{4}{5}\)

E = \(\frac{4}{3}+\frac{1}{4}+\frac{3}{4}\)

E = \(\frac{7}{3}\)

Vậy E = \(\frac{7}{3}\)

14 tháng 10 2016

Làm rõ ra đi bạn

31 tháng 8 2020

Bài 3 :

a) \(\left(\frac{1}{25}-0,6\right)^2:\frac{49}{125}+\left[\left(3\frac{1}{4}-6\frac{5}{9}\right)\cdot2\frac{2}{17}\right]\)

\(=\left(\frac{1}{25}-\frac{3}{5}\right)^2\cdot\frac{125}{49}+\left[\left(\frac{13}{4}-\frac{59}{9}\right)\cdot\frac{36}{17}\right]\)

\(=\left(-\frac{14}{25}\right)^2\cdot\frac{125}{49}+\left[\left(-\frac{119}{36}\right)\cdot\frac{36}{17}\right]\)

\(=-\frac{196}{625}\cdot\frac{125}{49}+\left(-7\right)=-\frac{4}{5}+\left(-7\right)=-\frac{39}{5}\)

31 tháng 8 2020

Trả lời :

\(\left(\frac{1}{25}-0,6\right)^2\div\frac{49}{125}+\left[\left(3\frac{1}{4}-6\frac{5}{9}\right)\times2\frac{2}{17}\right]\)

\(=\left(\frac{1}{25}-\frac{3}{5}\right)^2\div\frac{49}{125}+\left[\frac{-119}{36}\times\frac{36}{17}\right]\)

\(=\left(\frac{-14}{25}\right)^2\div\frac{49}{125}-7\)

\(=\frac{4}{5}-7\)

\(=\frac{-31}{5}\)

9 tháng 8 2016

a)\(25\frac{3}{5}:\left(\frac{-2}{3}\right)-15\frac{3}{5}:\left(\frac{-2}{3}\right)\)

\(=\left(25\frac{3}{5}-15\frac{3}{5}\right):\left(-\frac{2}{3}\right)\)

\(=10:\left(\frac{-2}{3}\right)\)

\(=-15\)

b)\(9.\left(\frac{-2}{3}\right)^3+\frac{1}{2}:5\)

\(=9.\frac{-8}{27}+\frac{1}{10}\)

\(=\frac{-8}{3}+\frac{1}{10}\)

\(=\frac{-77}{30}\)

c)\(\left[10\left(\frac{-1}{5}\right)^2+5\left(\frac{-1}{5}\right)+1\right]:\left(\frac{-1}{5}-1\right)\)

\(=\frac{2}{5}:\left(\frac{-6}{5}\right)\)

\(=\frac{-1}{3}\)

 

9 tháng 8 2016

\(a.25\frac{3}{5}:\left(-\frac{2}{3}\right)-15\frac{3}{5}:\left(-\frac{2}{3}\right)\)

\(=\frac{128}{5}:\left(-\frac{2}{3}\right)-\frac{75}{5}:\left(-\frac{2}{3}\right)\)

\(=\left(-\frac{192}{5}\right)-\left(-\frac{117}{5}\right)\)

\(=\frac{\left(-192\right)-\left(-117\right)}{5}\)

\(=-15\)

\(b.9.\left(-\frac{2}{3}\right)^3+\frac{1}{2}:5\)

\(=9.\left(-\frac{8}{27}\right)+\frac{1}{2}:5\)

\(=-\frac{8}{3}+\frac{1}{10}\)

\(=-\frac{77}{30}\)

\(c.\left[10\left(\frac{-1}{5}\right)^2+5\left(\frac{-1}{5}\right)+1\right]:\left(\frac{-1}{5}-1\right)\)

\(=\left[10\left(\frac{-1}{25}\right)+5\left(\frac{-1}{5}\right)+1\right]:\left(\frac{-1}{5}-1\right)\)

\(=\left[\frac{-2}{5}+\left(-1\right)+1\right]:\left(-\frac{6}{5}\right)\)

\(=\left(-\frac{2}{5}\right):\left(-\frac{6}{5}\right)\)

\(=\frac{1}{3}\)