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\(A=\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}\)
\(A=\frac{\left(2^2\right)^5.\left(3^2\right)^4-2.\left(2.3\right)^9}{2^{10}.3^8+\left(2.3\right)^8.2^2.5}\)
\(A=\frac{2^{10}.3^8-2.2^9.3^9}{2^{10}.3^8+2^8.3^8.2^2.5}\)
\(A=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}\)
\(A=\frac{2^{10}.3^8.\left(1-3\right)}{2^{10}.3^8.\left(1+5\right)}\)
\(A=\frac{2^{10}.3^8.\left(-2\right)}{2^{10}.3^8.6}\)
\(A=-\frac{2^{11}.3^8}{2^{10}.3^8.2.3}=-\frac{2^{11}.3^8}{2^{11}.3^9}=-\frac{1}{3}\)
Mình Gợi Ý Nè
Bạn đưa các cơ số ở cả tử và mẫu về cơ số 2 và 3 rồi đặt các thừa số chung ra ngoài và rút gọn
Mình Ra Kết Quả M = \(\frac{-1}{3}\)
Chúc bạn chăm ngoan học giỏi
và nhớ cho mình tích đậy !
\(\frac{\left(2^2\right)^5.\left(3^2\right)^4-2.6^8.2.3}{2^{10}.3^8+6^8.2^2.5}\)
\(\frac{2^{10}.3^8-2^2.6^8.3}{2^{10}.3^8+6^8.2^2.5}\)
Rồi sao nữa mình không biết
a: \(=\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+2^{10}\cdot3^8\cdot5}=\dfrac{2^{10}\cdot3^8\left(1-3\right)}{2^{10}\cdot3^8\left(1+5\right)}=\dfrac{-2}{6}=\dfrac{-1}{3}\)
b: \(=\dfrac{5^{16}\cdot3^{21}}{5^{15}\cdot3^{22}}=\dfrac{5}{3}\)
\(\frac{4^5\times9^4-2\times6^9}{2^{10}\times3^8+6^8\times20}=\frac{2^{10}\times3^8-2\times2^9\times3^9}{2^{10}\times3^8+2^8\times3^8\times2^2\times5}\)
\(=\frac{2^{10}\times3^8-2^{10}\times3^9}{2^{10}\times3^8+2^{10}\times3^8\times5}=\frac{2^{10}\times3^8\left(1-3\right)}{2^{10}\times3^8\left(1+5\right)}\)
\(=\frac{-2}{6}=\frac{-1}{3}\)
\(\dfrac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}=\dfrac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}=\dfrac{2^{10}.3^8\left(1-3\right)}{2^{10}.3^8\left(1-5\right)}=\dfrac{-2}{-4}=\dfrac{1}{2}\)
=\(\frac{2^{10}.3^8-2.2^9.3^9}{2^{10}.3^8+2^8.3^8.2^2.5}\)
=\(\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}\)
\(B=\dfrac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}=\dfrac{\left(2^2\right)^5.\left(3^2\right)^4-2.\left(6\right)^8.6}{2^{10}.3^8+\left(6\right)^8.20}=\dfrac{2^{10}.3^8-2\left(3.2\right)^8.6}{2^{10}.3^8+\left(3.2\right)^8.20}\)
\(B=\dfrac{2^{10}.3^8-2.3^8.2^8.6}{2^{10}.3^8+3^8.2^8.20}=\dfrac{3^8\left(2^{10}-2.2^8.6\right)}{3^8\left(2^{10}+2^8.20\right)}=\dfrac{2^{10}-2^9.6}{2^{10}+2^8.20}\)
\(B=\dfrac{2^8\left(2^2-2.6\right)}{2^8\left(2^2+20\right)}=\dfrac{2^2-2.6}{2^2+20}=\dfrac{4-12}{4+20}=\dfrac{-8}{24}=\dfrac{-1}{3}\)