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22 tháng 9 2019

\(\frac{\left(-0,25\right)^{-5}.9^4.\left(-2\right)^{-3}-2^{-2}.6^9}{2^9.3^6+6^6.40}\)

\(=\frac{\left(-4\right)^5.\left(3^2\right)^4.\left(-2\right)^{-3}-2^{-2}.\left(3.2\right)^9}{2^9.3^6+\left(2.3\right)^6.2^3.5}\)

\(=\frac{-\left(2^2\right)^5.3^8.\left(-2\right)^{-3}-2^{-2}.3^9.2^9}{2^9.3^6+2^6.3^6.2^3.5}\)

\(=\frac{-2^{10}3^8.\left(-2\right)^{-3}-2^7.3^9}{2^9.3^6+2^9.3^6.5}\)

\(=\frac{2^73^8.-2^7.3^9}{2^9.3^6+2^9.3^6.5}\)

\(=\frac{2^7.3^8.\left(1-3\right)}{2^9.3^6.\left(1+5\right)}\)

\(=\frac{3^2.\left(-2\right)}{2^2.6}\)

\(=\frac{-3}{4}\)

16 tháng 2 2023

\(C=\dfrac{5\times2^{12}\times3^8-3^9\times2^{12}}{2^2\times2^{13}\times3^8+2\times2^{12}\times\left(-3^9\right)}=\dfrac{3^8\times2^{12}\times\left(5-3\right)}{2^{15}\times3^8+2^{13}\times\left(-3\right)^9}\)

\(=\dfrac{3^8\times2^{12}\times2}{2^{13}\times3^8\times\left(4-3\right)}=\dfrac{1}{1}=1\)

\(#PaooNqoccc\)

1 tháng 8 2018

a. \(\frac{20^5.5^{10}}{100^5}\)

\(=\frac{20^5.\left(5^2\right)^5}{100^5}\)

\(=\frac{20^5.25^5}{100^5}\)

\(=\frac{500^5}{100^5}\)

\(=\left(\frac{500}{100}\right)^5\)

\(=5^5=3125\)

b. \(\frac{\left(0,9\right)^5}{\left(0,3\right)^6}\)

\(=\frac{\left(0,9\right)^5}{\left(0,3\right)^5.0,3}\)

\(=\left(\frac{0,9}{0,3}\right)^5.\frac{1}{0,3}\)

\(=3^5.\frac{1}{0,3}\)

\(=810\)

c. \(\frac{6^3+3.6^2+3^3}{-13}\)

\(=\frac{\left(3.2\right)^3+3.\left(3.2\right)^2+3^3}{-13}\)

\(=\frac{3^3\left(2^3+2^2+1\right)}{-13}\)

\(=\frac{3^3.13}{-13}\)

\(=\left(-3\right)^3\)

\(=-27\)

26 tháng 7 2017

a, \(4\times\left(-\dfrac{1}{2}\right)^3-2\times\left(-\dfrac{1}{2}\right)^2+3\times\left(-\dfrac{1}{2}\right)+1\)

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

\(=\left(-\dfrac{1}{2}\right)\left(-2+1+3\right)+1\)

\(=\left(-\dfrac{1}{2}\right)2+1\)

\(=-1+1\)

\(=0\)

@Trịnh Thị Thảo Nhi

29 tháng 4 2018

a, 4×(−12)3−2×(−12)2+3×(−12)+14×(−12)3−2×(−12)2+3×(−12)+1

=(−12)[(4×−12)−(2×−12)+3]+1=(−12)[(4×−12)−(2×−12)+3]+1

=(−12)(−2+1+3)+1=(−12)(−2+1+3)+1

=(−12)2+1=(−12)2+1

=−1+1=−1+1

=0=0