BN

Giúp vsloading... ạ

NT
1 tháng 2 2024 lúc 19:49

Bài 6:

\(A=\dfrac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\cdot3\right)^6}-\dfrac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\)

\(=\dfrac{2^{12}\cdot3^5-2^{12}\cdot3^4}{2^{12}\cdot3^6}-\dfrac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot7^3\cdot2^3}\)

\(=\dfrac{2^{12}\cdot3^4\left(3-1\right)}{2^{12}\cdot3^6}-\dfrac{5^{10}\cdot7^3\left(1-7\right)}{5^9\cdot7^3\left(1+2^3\right)}\)

\(=\dfrac{-2}{3^2}+\dfrac{5\cdot6}{9}=\dfrac{-2}{9}+\dfrac{30}{9}=\dfrac{28}{9}\)

Bài 5:

a: \(A=\dfrac{2^{19}\cdot27^3\cdot5-15\cdot\left(-4\right)^9\cdot9^4}{6^9\cdot2^{10}-\left(-12\right)^{10}}\)

\(=\dfrac{2^{19}\cdot3^9\cdot5+5\cdot3^9\cdot2^{18}}{2^{19}\cdot3^9-2^{20}\cdot3^{10}}\)

\(=\dfrac{2^{18}\cdot3^9\cdot5\left(2+1\right)}{2^{19}\cdot3^9\left(1-2\cdot3\right)}\)

\(=\dfrac{1}{2}\cdot\dfrac{5\cdot3}{-5}=\dfrac{-3}{2}\)

b: \(B=\dfrac{8^5\left(-5\right)^8+\left(-2\right)^5\cdot10^9}{2^{16}\cdot5^7+20^8}\)

\(=\dfrac{2^{15}\cdot5^8-2^{14}\cdot5^9}{2^{16}\cdot5^7+2^{16}\cdot5^8}\)

\(=\dfrac{2^{14}\cdot5^8\left(2-5\right)}{2^{16}\cdot5^7\left(1+5\right)}=\dfrac{5}{4}\cdot\dfrac{-3}{6}=\dfrac{5}{4}\cdot\dfrac{-1}{2}=-\dfrac{5}{8}\)

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