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NN
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10 tháng 9 2019
\(\frac{-3^{10}\times15^7}{25^4\times\left(-9\right)^9}=\frac{\left[\left(-1\right).3\right]^{10}\times\left(3.5\right)^7}{\left(5^2\right)^4\times\left[\left(-1\right).3^2\right]^9}=\frac{\left(-1\right)^{10}.3^{10}.3^7.5^7}{5^8.\left(-1\right)^9.\left(3^2\right)^9}=\frac{1.3^{17}.5^7}{5^8.\left(-1\right).3^{18}}\)
\(=\frac{1}{5.\left(-1\right).3}=\frac{1}{-15}=\frac{-1}{15}\)
10 tháng 9 2019
\(\frac{-3^{10}x15^7}{25^4.\left(-9\right)^9}=\frac{-3^{10}.\left(5.3\right)^7}{\left(5.5\right)^4.\left[-\left(3^2\right)^9\right]}=\frac{-3^{10}.5^7.3^7}{5^4.5^4.\left[-\left(3^{18}\right)\right]}=\frac{3^7}{5.3^8}=\frac{3}{5}\)
17 tháng 9 2020
a) \(4\frac{5}{9}:\left(-\frac{5}{7}\right)+\frac{49}{9}:\left(-\frac{5}{7}\right)=\frac{41}{9}:\left(-\frac{5}{7}\right)+\frac{49}{9}:\left(-\frac{5}{7}\right)\)
\(=\frac{41}{9}\cdot\left(-\frac{7}{5}\right)+\frac{49}{9}\cdot\left(-\frac{7}{5}\right)=\left(\frac{41}{9}+\frac{49}{9}\right)\cdot\left(-\frac{7}{5}\right)=10\cdot\left(-\frac{7}{5}\right)=-14\)
b) \(\left(\frac{-3}{5}+\frac{4}{9}\right):\frac{7}{11}+\left(\frac{-2}{5}+\frac{5}{9}\right):\frac{7}{11}\)
\(=\left(\frac{-3}{5}+\frac{4}{9}+\frac{-2}{5}+\frac{5}{9}\right):\frac{7}{11}\)
\(=\left(\frac{-3}{5}+\frac{-2}{5}+\frac{4}{9}+\frac{5}{9}\right):\frac{7}{11}\)
\(=\left(-1+1\right):\frac{7}{11}=0\cdot\frac{11}{7}=0\)
c) \(\left(\frac{3}{4}\right)^4\cdot\left(\frac{8}{9}\right)^2=\left(\frac{3}{4}\right)^2\cdot\left(\frac{3}{4}\right)^2\cdot\left(\frac{8}{9}\right)^2=\left(\frac{3}{4}\cdot\frac{3}{4}\cdot\frac{8}{9}\right)^2\)
\(=\left(\frac{1}{2}\right)^2=\frac{1}{4}\)
d) \(\left(-\frac{3}{5}\right)^6\cdot\left(-\frac{5}{3}\right)^5=\left(-\frac{3}{5}\right)^5\cdot\left(-\frac{3}{5}\right)\cdot\left(-\frac{5}{3}\right)^5=\left[\left(-\frac{3}{5}\right)\cdot\left(-\frac{5}{3}\right)\right]^5\cdot\left(-\frac{3}{5}\right)\)
\(=1^5\cdot\left(-\frac{3}{5}\right)=1\cdot\left(-\frac{3}{5}\right)=-\frac{3}{5}\)
e) \(\frac{8^{14}}{4^4\cdot64^5}=\frac{\left(2^3\right)^{14}}{\left(2^2\right)^4\cdot\left(2^6\right)^5}=\frac{2^{42}}{2^8\cdot2^{30}}=\frac{2^{42}}{2^{38}}=2^4=16\)
f) \(\frac{9^{10}\cdot27^7}{81^7\cdot3^{15}}=\frac{\left(3^2\right)^{10}\cdot\left(3^3\right)^7}{\left(3^4\right)^7\cdot3^{15}}=\frac{3^{20}\cdot3^{21}}{3^{28}\cdot3^{15}}=\frac{3^{41}}{3^{43}}=3^{-2}=\frac{1}{3^2}=\frac{1}{9}\)
14 tháng 5 2016
Ta có: F= (100-12) (100-22)...(100-252)
=> F= (100-12)...(100-102)...(100-252)
=> F= (100-12)...0...(100-252)
=> F= 0
Vậy F= 0
\(\frac{\left(-3\right)^{10}\cdot15^5}{25^3\cdot\left(-9\right)^7}=\frac{3^{15}\cdot5^5}{-5^6\cdot3^{14}}=\frac{-3}{5}\)