\(B=\frac{\left(5^4\right)-\left(5^3\right)}{125}\)

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20 tháng 5 2016

Rễ thui ta có: \(B=\frac{\left(5^4\right)-\left(5^3\right)}{125}=\frac{500}{125}=4\)

5= 5*5*5*5= 625 

53=5*5*5=125 

54 - 5= 625 - 125 = 500 

Vậy B = 500/ 125 

vậy B = 4 

6 tháng 11 2017

a, \(\dfrac{2\cdot8^4\cdot27^2+4\cdot6^9}{2^7\cdot6^7+2^7\cdot40\cdot9^4}\)

=\(\dfrac{2\cdot\left(2^3\right)^4\cdot\left(3^3\right)^2+2^2\cdot2^9\cdot3^9}{2^7\cdot2^7\cdot3^7+2^7\cdot2^3\cdot5\cdot\left(3^2\right)^4}\)

=\(\dfrac{2\cdot2^{12}\cdot3^6+2^{11}\cdot3^9}{2^{14}\cdot3^7+2^{10}\cdot5\cdot3^8}\)

=\(\dfrac{2^{11}\cdot3^6\cdot\left(2^2+3^3\right)}{2^{10}\cdot3^7\cdot\left(2^4+5\cdot3\right)}\)

=\(\dfrac{2^{11}\cdot3^6\cdot31}{2^{10}\cdot3^7\cdot31}\)

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

6 tháng 11 2017

b, \(\dfrac{\dfrac{8}{27}\cdot\dfrac{9}{16}\cdot\left(-1\right)}{\dfrac{4}{25}\cdot\dfrac{-125}{1728}}\)

=\(\dfrac{\dfrac{8\cdot9\cdot\left(-1\right)}{27\cdot16}}{\dfrac{4\cdot\left(-125\right)}{25\cdot1728}}\)

=\(\dfrac{\dfrac{-1}{6}}{\dfrac{-5}{432}}\)

=\(\dfrac{-1}{6}\cdot\dfrac{-432}{5}\)

=\(\dfrac{72}{5}\)

1 tháng 11 2019

2a) \(\frac{3^6+45^4-15^3.4^5}{27^4.25^3+45^6}\)

\(\frac{3^6+\left(3^2.5\right)^4-\left(3.5\right)^3.\left(2^2\right)^5}{\left(3^3\right)^4.\left(5^2\right)^3+\left(3^2.5\right)^6}\)

\(\frac{3^6+3^8.5^4-3^3.5^3.4^{10}}{3^{12}.5^6-3^{12}.5^6}=\frac{3^3.\left(3^3+3^5.5^4-5^3.4^{10}\right)}{0}\)(xem lại đề)

b)  \(\frac{\left(\frac{2}{5}\right)^7.5^7+\left(\frac{16}{3}\right)^3:\left(\frac{4}{9}\right)^3}{2^7.5^2+512}\)

\(\frac{\left(\frac{2}{5}.5\right)^7+\left(\frac{16}{3}:\frac{4}{9}\right)^3}{2^7.5^2+2^9}\)

\(\frac{2^7+12^3}{2^7\left(5^2+2^2\right)}\)

\(\frac{2^7+\left(2^2.3\right)^3}{2^7.29}\)

\(\frac{2^7+2^6.3^3}{2^7.29}\)

\(\frac{2^6\left(1+27\right)}{2^7.29}=\frac{28}{2.29}=\frac{14}{29}\)

1 tháng 11 2019

mk xin lỗi bn nhé...mk vt nhầm đề...mk vt lại nha:

2a,\(\frac{3^6+45^4-15^3.9^5}{27^4.25^3+45^6}\)

NM
12 tháng 10 2021

ta có:

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12 tháng 10 2021

Thầy làm chi tiết giúp em được ko ạ ?

13 tháng 9 2019

\(B=\frac{\left[\frac{2}{3}\right]^3\cdot\left[-\frac{3}{4}\right]^2\cdot\left[-1\right]^5}{\left[\frac{2}{5}\right]^2\cdot\left[-\frac{5}{12}\right]^3}\)

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

\(=\frac{\frac{8}{27}\cdot\frac{9}{16}\cdot\left[-1\right]}{\frac{4}{25}\cdot\frac{-125}{\left[2^2\cdot3\right]^3}}\)

\(=\frac{\frac{1}{3}\cdot\frac{1}{2}\cdot\left[-1\right]}{\frac{4}{25}\cdot\frac{-125}{\left[2^2\right]^3\cdot3^3}}\)

\(=\frac{\frac{1\cdot1\cdot\left[-1\right]}{3\cdot2\cdot1}}{\frac{4}{25}\cdot\frac{-125}{4^3\cdot3^3}}\)

\(=\frac{\frac{-1}{6}}{\frac{4}{25}\cdot\frac{-125}{64\cdot27}}=\frac{\frac{-1}{6}}{\frac{4}{1}\cdot\frac{-5}{64\cdot27}}\)

\(=\frac{\frac{-1}{6}}{4\cdot\frac{-5}{64\cdot27}}=\frac{\frac{-1}{6}}{-\frac{20}{64\cdot27}}=\frac{72}{5}\)