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

Đề bài yêu cầu gì?

a) \(\dfrac{27^3\cdot11+9^5\cdot5}{3^9\cdot2^4}\)

\(=\dfrac{3^9\cdot11+3^{10}\cdot5}{3^9\cdot2^4}\)

\(=\dfrac{3^9\cdot\left(11+3\cdot5\right)}{3^9\cdot2^4}\)

\(=\dfrac{11+15}{16}\)

\(=\dfrac{26}{16}\)

\(=\dfrac{13}{8}\)

b) \(\dfrac{5^8+2^2\cdot25^4+2^3\cdot125^3-15^4\cdot5^4}{4^2\cdot625^2}\)

\(=\dfrac{5^8+2^2\cdot5^8+2^3\cdot5^9-3^4\cdot5^4\cdot5^4}{2^4\cdot5^8}\)

\(=\dfrac{5^8\cdot\left(1+2^2+2^3\cdot5-3^4\right)}{5^8\cdot2^4}\)

\(=\dfrac{1+4+40-81}{16}\)

\(=\dfrac{-36}{16}\)

\(=\dfrac{-9}{4}\)

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

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

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

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

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

\(A=\frac{4^5.9^4-2.6^9}{2^{10}.3^8-6^8.20}\)

\(A=\frac{\left(2^2\right)^5.\left(3^2\right)^4-2.\left(2.3\right)^9}{2^{10}.3^8-\left(2.3\right)^8.2^2.5}\)

\(A=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8-2^{10}.3^8.5}\)

\(A=\frac{2^{10}.\left(3^8-3^9\right)}{2^{10}.3^8.\left(1-5\right)}=\frac{3^8-3^9}{3^8.\left(-4\right)}=\frac{3^8.\left(1-3\right)}{3^8.\left(-4\right)}=\frac{-2}{-4}=\frac{1}{2}\)

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 = \(\frac{1}{2}\)

1: \(=5^{20}\cdot\left(\dfrac{1}{5}\right)^{20}+\left(\dfrac{-3}{4}\cdot\dfrac{-4}{3}\right)^8-1\)

=1+1-1=1

2: \(=\dfrac{15-8}{6}\cdot\dfrac{6}{7}+\left(-\dfrac{3}{2}\right)^2\)

=1+9/4

=13/4

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

\(=\dfrac{2^{10}\cdot3^8\left(1-3\right)}{3^8\cdot2^{10}\cdot6}=\dfrac{-2}{6}=\dfrac{-1}{3}\)

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

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

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

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

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

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

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

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

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

a: \(=6-\dfrac{2}{3}+\dfrac{1}{2}-5-\dfrac{5}{3}+\dfrac{3}{2}-3+\dfrac{7}{3}-\dfrac{5}{2}\)

\(=\left(6-5-3\right)+\left(-\dfrac{2}{3}-\dfrac{5}{3}+\dfrac{7}{3}\right)+\left(\dfrac{1}{2}+\dfrac{3}{2}-\dfrac{5}{2}\right)\)

\(=-2-\dfrac{1}{2}=-\dfrac{5}{2}\)

b: \(=\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+2^8\cdot3^8\cdot2^2\cdot5}=\dfrac{2^{10}\cdot3^8\cdot\left(-2\right)}{2^{10}\cdot3^8\left(1+5\right)}=\dfrac{-2}{6}=-\dfrac{1}{3}\)

Ta có

\(E=\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}\cdot\frac{5^4.20^4}{25^5.4^5}\)

\(=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}\cdot\frac{2^8.5^8}{5^{10}.2^{10}}\)

\(=\frac{2^{10}.3^8.\left(1-3\right)}{2^{10}.3^8.\left(1+5\right)}\cdot\frac{1}{5^2.2^2}\)

\(=\frac{\left(-2\right)}{6}\cdot\frac{1}{100}=-\frac{1}{3}\cdot\frac{1}{100}=-\frac{1}{300}\)

Vậy : \(E=-\frac{1}{300}\)

Bài làm

\(E=\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}.\frac{5^4.20^4}{25^5.4^5}\)

\(\Rightarrow E=\frac{2^{10}.3^8-2.2^9.3^9}{2^{10}.3^8+2^8.3^8.2^2.5}.\frac{5^4.4^4.5^4}{5^{10}.4^5}\)

\(\Rightarrow E=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}.\frac{5^8.4^4}{5^{10}.4^5}\)

\(\Rightarrow E=\frac{2^{10}\left(3^8-3^9\right)}{2^{10}\left(3^8+3^8.5\right)}.\frac{1}{5^2.4}\)

\(\Rightarrow E=\frac{3^8-3^9}{3^8\left(1+5\right)}.\frac{1}{100}\)

\(\Rightarrow E=\frac{3^8\left(1-3\right)}{3^8\left(1+5\right)}.\frac{1}{100}\)

\(\Rightarrow E=-\frac{2}{6}.\frac{1}{100}\)

\(\Rightarrow E=-\frac{1}{300}\)

a)

\(5A=5+5^2+.....+5^{101}\)

\(\Rightarrow5A-A=\left(5+5^2+.....+5^{101}\right)-\left(1+5+.....+5^{100}\right)\)

\(\Rightarrow4A=5^{101}-1\)

\(\Rightarrow A=\frac{5^{101}-1}{4}\)

b)

\(2B=1+\left(\frac{1}{2}\right)^2+....+\left(\frac{1}{2}\right)^{100}\)

\(\Rightarrow2B-B=\left(1+\frac{1}{2^2}+.....+\frac{1}{2^{100}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+......+\frac{1}{2^{99}}\right)\)

\(\Rightarrow B=1-\frac{1}{2^{100}}\)