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: \(=\dfrac{5^{16}\cdot3^{21}}{5^{15}\cdot3^{22}}=\dfrac{5}{3}\)

cho xin đáp án chi tiết điiiii

^{nhanh hộ mình cái}

\(\dfrac{\left(-0.25\right)^{-5}\cdot9^4\cdot\left(-2\right)^{-3}-2^{-2}\cdot6^9}{2^9\cdot3^6+6^6\cdot40}\)

\(=\dfrac{2^7\cdot3^8-2^8\cdot3^9}{2^9\cdot3^6+2^9\cdot3^6\cdot5}\)

\(=\dfrac{2^7\cdot3^8\cdot\left(-5\right)}{2^9\cdot3^6\cdot6}\)

\(=\dfrac{1}{4}\cdot9\cdot\dfrac{-5}{6}=\dfrac{-45}{24}=\dfrac{-15}{8}\)

\(A=\dfrac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\)

\(=\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+3^8\cdot2^{10}\cdot5}\)

\(=\dfrac{2^{10}\cdot3^8\left(1-3\right)}{2^{10}\cdot3^8\cdot\left(1+5\right)}\)

\(=-\dfrac{2}{6}=-\dfrac{1}{3}\)

\(A=\dfrac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}\)
\(=\dfrac{2^{10}.3^8-2.2^9.3^9}{2^{10}.3^8+2^8.3^8.2^2.5}\)
\(=\dfrac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}\)
\(=\dfrac{2^{10}.3^8\left(1-3\right)}{2^{10}.3^8\left(1+5\right)}\)
\(=-\dfrac{2}{6}=-\dfrac{1}{3}\)

\(=\dfrac{\left(2^2\right)^5.\left(3^2\right)^4-2.3^9.2^9}{2^{10}.3^8+2^8.3^8.2^2.5}=\dfrac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}=\dfrac{2^{10}.3^8\left(1-3\right)}{2^{10}.3^8\left(1+5\right)}=\dfrac{-2}{6}=-\dfrac{1}{3}\)

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= \(\dfrac{4^5.9^4-2.6^9}{2^{10}.3+6^8.20}\)

= \(\dfrac{\left(2^2\right)^5.\left(3^3\right)^4-2.6^9}{2^{10}.3+6^8.20}\)

=\(\dfrac{2^{10}.3^{12}-2.6^9}{2^{10}.3+6^8.20}\)

= \(\dfrac{3^{11}-6}{10}\)

\(A=1+5+5^2+5^3+5^4+...+5^{49}+5^{50}\)

\(\Rightarrow5A=5\left(1+5+5^2+5^3+5^4+...+5^{49}+5^{50}\right)\)

\(\Rightarrow5A=5+5^2+5^3+5^4+5^5+...+5^{50}+5^{51}\)

\(\Rightarrow5A-A=\left(5+5^2+5^3+5^4+5^5+...+5^{50}+5^{51}\right)-\left(1+5+5^2+5^3+5^4+...+5^{49}+5^{50}\right)\)\(\Rightarrow4A=5^{51}-1\)

\(\Rightarrow A=\dfrac{5^{51}-1}{4}\)

\(\dfrac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}=\dfrac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}=\dfrac{2^{10}.3^8\left(1-3\right)}{2^{10}.3^8\left(1-5\right)}=\dfrac{-2}{-4}=\dfrac{1}{2}\)

\(\dfrac{\left(-0.25\right)^{-5}\cdot9^4\left(-2\right)^{-3}-2^{-3}\cdot6^9}{2^9\cdot3^6+6^6\cdot40}\)

\(=\dfrac{2^7\cdot3^8-2^6\cdot3^9}{2^9\cdot3^6+3^6\cdot2^9\cdot5}\)

\(=\dfrac{2^6\cdot3^8\left(2-3\right)}{2^9\cdot3^6\cdot6}\)

\(=\dfrac{1}{2^3}\cdot3^2\cdot\dfrac{-1}{6}\)

\(=\dfrac{-9}{6\cdot2^3}=\dfrac{-3}{2^4}=\dfrac{-3}{16}\)

\(\dfrac{4^5\cdot9^4}{8^3\cdot27^3}=\dfrac{\left(2^2\right)^5\cdot\left(3^2\right)^4}{\left(2^3\right)^3\cdot\left(3^3\right)^3}=\dfrac{2^{10}\cdot3^8}{2^9\cdot3^9}=\dfrac{2}{3}\)

\(\dfrac{4^{20}\cdot3^{35}}{2^{37}\cdot27^{12}}=\dfrac{\left(2^2\right)^{20}\cdot3^{35}}{2^{37}\cdot\left(3^3\right)^{12}}=\dfrac{2^{40}\cdot3^{35}}{2^{37}\cdot3^{36}}=\dfrac{2^3}{3}\)

\(\dfrac{5^4\cdot20^4}{25^5\cdot4^5}=\dfrac{5^4\cdot5^4\cdot4^4}{5^5\cdot5^5\cdot4^5}=\dfrac{1}{5^2\cdot4}=\dfrac{1}{100}\)

\(\dfrac{2^{15}\cdot9^4}{6^6\cdot8^3}=\dfrac{2^{15}\cdot\left(3^2\right)^4}{2^6\cdot3^6\cdot\left(2^3\right)^3}=\dfrac{2^{15}\cdot3^8}{2^6\cdot3^6\cdot2^9}=3^2\)