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NT
3 tháng 1 2018
\(M=-\left(\dfrac{4}{1.5}+\dfrac{4}{5.9}+...+\dfrac{4}{n\left(n+4\right)}\right)\\ =-\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{n}-\dfrac{1}{n+4}\right)\\ =-\left(1-\dfrac{1}{n+4}\right)\\ =-\left(\dfrac{n+3}{n+4}\right)\)
NT
3 tháng 1 2018
\(-4x\left(x-5\right)-2x\left(8-2x\right)=-3\\ \Rightarrow-4x^2+20x-16x+4x^2=-3\\ \Rightarrow4x=-3\\ \Rightarrow x=-\dfrac{3}{4}\)
SS
25 tháng 6 2019
\(A=\frac{4^5.9^4-2.6^9}{2^{10}.3^8-6^8.20}\)
\(A=\frac{\left(2^2\right)^5.\left(3^2\right)^4-2.\left(2.3\right)^9}{2^{10}.3^8-\left(2.3\right)^8.2^2.5}\)
\(A=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8-2^{10}.3^8.5}\)
\(A=\frac{2^{10}.\left(3^8-3^9\right)}{2^{10}.3^8.\left(1-5\right)}=\frac{3^8-3^9}{3^8.\left(-4\right)}=\frac{3^8.\left(1-3\right)}{3^8.\left(-4\right)}=\frac{-2}{-4}=\frac{1}{2}\)
Vậy A = \(\frac{1}{2}\)
\(B=\frac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}\)
\(B=\frac{2^{19}.\left(3^3\right)^3+3.5.\left(2^2\right)^9.\left(3^2\right)^4}{\left(2.3\right)^9.2^{10}+\left(2^2.3\right)^{10}}\)
\(B=\frac{2^{19}.3^9+3.5.2^{18}.3^8}{2^9.3^9.2^{10}+2^{20}.3^{10}}\)
\(B=\frac{2^{19}.3^9+3^9.2^{18}.5}{2^{19}.3^9+2^{20}.3^{10}}\)
\(B=\frac{2^{18}.3^9.\left(2+5\right)}{2^{19}.3^9\left(1+2.3\right)}=\frac{7}{2.7}=\frac{1}{2}\)
Vậy B = \(\frac{1}{2}\)
20 tháng 12 2022
a: \(=\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+2^{10}\cdot3^8\cdot5}=\dfrac{2^{10}\cdot3^8\left(1-3\right)}{2^{10}\cdot3^8\left(1+5\right)}=\dfrac{-2}{6}=\dfrac{-1}{3}\)
b: \(=\dfrac{5^{16}\cdot3^{21}}{5^{15}\cdot3^{22}}=\dfrac{5}{3}\)
PE
0
NN
1
2 tháng 10 2019
\(2^5\).\(9^5\).\(2^8\).\(9^8\)
=(\(2^5\).\(2^8\)).(\(9^5\).\(9^8\))
=\(^{2^{13}}\).\(9^{13}\)
=\(^{2.9^{13}}\)
=\(18^{13}\)
3 tháng 11 2015
\(2^5.9^5.2^8.9^8=\left(2^5.2^8\right).\left(9^5.9^8\right)=2^{13}.9^{13}=\left(2.9\right)^{13}=18^{13}\)
\(2x=5.9^2:8\)
\(2x=5.81:8\)
\(16x=405\)
\(x=\dfrac{405}{16}\)
\(5\cdot9^2:8=2x\\\Rightarrow 2x=5\cdot81:8\\\Rightarrow2x=405:8\\\Rightarrow2x=\dfrac{405}{8}\\\Rightarrow x=\dfrac{405}{16}\\Vậy:x=\dfrac{405}{16}\)