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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: \(=\dfrac{5^{16}\cdot3^{21}}{5^{15}\cdot3^{22}}=\dfrac{5}{3}\)
\(A=\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}\)
\(A=\frac{\left(2^2\right)^5.\left(3^2\right)^4-2.\left(2.3\right)^9}{2^{10}.3^8+\left(2.3\right)^8.2^2.5}\)
\(A=\frac{2^{10}.3^8-2.2^9.3^9}{2^{10}.3^8+2^8.3^8.2^2.5}\)
\(A=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}\)
\(A=\frac{2^{10}.3^8.\left(1-3\right)}{2^{10}.3^8.\left(1+5\right)}\)
\(A=\frac{2^{10}.3^8.\left(-2\right)}{2^{10}.3^8.6}\)
\(A=-\frac{2^{11}.3^8}{2^{10}.3^8.2.3}=-\frac{2^{11}.3^8}{2^{11}.3^9}=-\frac{1}{3}\)
=\(\frac{2^{10}.3^8-2.2^9.3^9}{2^{10}.3^8+2^8.3^8.2^2.5}\)
=\(\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}\)
\(B=\dfrac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}=\dfrac{\left(2^2\right)^5.\left(3^2\right)^4-2.\left(6\right)^8.6}{2^{10}.3^8+\left(6\right)^8.20}=\dfrac{2^{10}.3^8-2\left(3.2\right)^8.6}{2^{10}.3^8+\left(3.2\right)^8.20}\)
\(B=\dfrac{2^{10}.3^8-2.3^8.2^8.6}{2^{10}.3^8+3^8.2^8.20}=\dfrac{3^8\left(2^{10}-2.2^8.6\right)}{3^8\left(2^{10}+2^8.20\right)}=\dfrac{2^{10}-2^9.6}{2^{10}+2^8.20}\)
\(B=\dfrac{2^8\left(2^2-2.6\right)}{2^8\left(2^2+20\right)}=\dfrac{2^2-2.6}{2^2+20}=\dfrac{4-12}{4+20}=\dfrac{-8}{24}=\dfrac{-1}{3}\)
\(A=\dfrac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\)
\(=\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+3^8\cdot2^{10}\cdot5}\)
\(=\dfrac{2^{10}\cdot3^8\left(1-3\right)}{2^{10}\cdot3^8\cdot\left(1+5\right)}\)
\(=-\dfrac{2}{6}=-\dfrac{1}{3}\)
\(A=\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}\)
\(=\frac{2^{10}.3^8-2.2^9.3^9}{2^{10}.3^8+2^8.3^8.2^2.5}\)
\(=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.2^8+2^{10}.3^8.5}\)
\(=\frac{2^{10}.3^8\left(1-3\right)}{2^{10}.3^8\left(1+5\right)}\)
\(=\frac{-2}{6}=-\frac{1}{3}\)
\(A=\frac{\left(2^2\right)^5.\left(3^2\right)^4-2.2^9.3^9}{2^{10}.3^8+2^8.3^8.2^2.5}\)
\(A=\frac{2^{10}.3^8-2^{11}.3^9}{2^{10}.3^8+2^{10}.3^8.5}\)
\(A=\frac{2^{10}.3^8.\left(1-2.3\right)}{2^{10}.3^8.\left(1+5\right)}\)
\(A=\frac{1-6}{6}\)
\(A=\frac{-5}{6}\)
\(\frac{\left(2^2\right)^5.\left(3^2\right)^4-2.6^8.2.3}{2^{10}.3^8+6^8.2^2.5}\)
\(\frac{2^{10}.3^8-2^2.6^8.3}{2^{10}.3^8+6^8.2^2.5}\)
Rồi sao nữa mình không biết
\(A=\frac{\left(2^2\right)^5.\left(3^2\right)^4-2.\left(2.3\right)^9}{2^{10}.3^8+\left(2.3\right)^8.2^2.5}=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}=\frac{2^{10}.3^8\left(1-3\right)}{2^{10}.3^8\left(1+5\right)}=\frac{-2}{6}=\frac{-1}{3}\)
A = \(\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}=\frac{\left(2^2\right)^5.\left(3^2\right)^4-2.\left(2.3\right)^9}{2^{10}.3^8+\left(2.3\right)^8.2^2.5}=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}\)\(=\frac{2^{10}.3^8.\left(1-3\right)}{2^{10}.3^8.\left(1+5\right)}=\frac{-2}{6}=\frac{-1}{3}\)
a, ta có A.5 = 5 ( 1+5 +52 +...........+549 +550)
5A = 5 +52 +53 +............... + 550 +551
5A-A = (5 +52 +53 +............+ 551) - (1+5+52 +......+550)
4A = 551 -1
A =\(\dfrac{5^{51}-1}{4}\)
vậy A =
b, B= \(\dfrac{4^5.9^4-2.6^9}{2^{10}.3+6^8.20}\)
= \(\dfrac{\left(2^2\right)^5.\left(3^3\right)^4-2.6^9}{2^{10}.3+6^8.20}\)
=\(\dfrac{2^{10}.3^{12}-2.6^9}{2^{10}.3+6^8.20}\)
= \(\dfrac{3^{11}-6}{10}\)