Đề bài yêu cầu gì?

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

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

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

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

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

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

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

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

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

Bạn dùng thanh công cụ này để ghi lại đề nhé! Đề ghi vậy đọc hơi khó hiểu tí :))

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

tự làm cho giỏi

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

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

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

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

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

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 = \(\frac{1}{2}\)

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

                                   \(=\frac{2^{10}.3^8\left(1-3\right)}{2^{10}.3^8.\left(1+5\right)}=-\frac{2}{6}=-\frac{1}{3}\)

\(B=\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}=\frac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{2^{12}.3^{12}-2^{11}.3^{11}}=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}-2^{11}.3^{11}}=\frac{2^{11}.2.3^{10}.\left(1+5\right)}{2^{11}.3^{10}.3.\left(6-1\right)}=\frac{12}{15}=\frac{4}{5}\)

\(A=\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}=\frac{2^{10}.3^8-2^2.3^9}{2^{10}.3^8-\left(-\left(2^2.3^8.5\right)\right)}=\frac{2^2.3^9}{-\left(2^2.3^8.5\right)}=-\frac{3}{5}\)

thanks bn Phạm Ngọc Thạch nhiều