Ta có: \(\frac{1}{2^2}<\frac{1}{1\cdot2}=1-\frac12\)

\(\frac{1}{3^2}<\frac{1}{2\cdot3}=\frac12-\frac13\)

...

\(\frac{1}{9^2}<\frac{1}{8\cdot9}=\frac18-\frac19\)

Do đó: \(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{9^2}<1-\frac12+\frac12-\frac13+\cdots+\frac18-\frac19\)

=>\(A<1-\frac19\)

=>A<8/9(1)

Ta có: \(\frac{1}{2^2}>\frac{1}{2\cdot3}=\frac12-\frac13\)

\(\frac{1}{3^2}>\frac{1}{3\cdot4}=\frac13-\frac14\)

...

\(\frac{1}{9^2}>\frac{1}{9\cdot10}=\frac19-\frac{1}{10}\)

Do đó: \(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{10^2}>\frac12-\frac13+\frac13-\frac14+\cdots+\frac19-\frac{1}{10}\)

=>A>1/2-1/5

=>A>2/5(2)

Từ (1),(2) suy ra 2/5<A<8/9

A<1-1/2+1/2-1/3+...+1/8-1/9=1-1/9=8/9

A>1/2-1/3+1/3-1/4+...+1/9-1/10=1/2-1/10=2/5

=>2/5<A<8/9

\(A=3+\frac{3}{2}+\frac{3}{2^2}+\frac{3}{2^3}+..+\frac{3}{2^9}\)

\(2A=6+3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^8}\)

\(2A-A=\left[6+3+\frac{3}{2}+...+\frac{3}{2^8}\right]-\left[3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}\right]\)

=> A = 6 + 3/2^9 = 3075/512

Mik ghi nhầm, thay A bằng S giùm nha

Cảm ơn

Học giỏi nhé pạn

mong bn lan sau vt de hiu hon!please

Đề bài yêu cầu j vậy?

mũ 2 tui ko bt

a) \(\frac{75^3.3^7}{81^4.5^6}=\frac{5^3.3^3.5^3.3^7}{\left(3^4\right)^4.5^6}=\frac{5^6.3^3.3^7}{3^{16}.5^6}=\frac{3^{10}}{3^{16}}=\frac{1}{3^6}=\frac{1}{729}\)
b) \(\frac{6^6.4^2}{3^{12}.2^8}=\frac{2^6.3^6.\left(2^2\right)^2}{3^{12}.2^8}=\frac{2^6.3^6.2^4}{3^{12}.2^8}=\frac{2^{10}.3^6}{3^{12}.2^8}=\frac{2^2.1}{3^6}=\frac{4}{729}\)
c) \(\frac{34^5.2^5}{2^{14}.17^5}=\frac{2^5.17^5.2^5}{2^{14}.17^5}=\frac{2^{10}}{2^{14}}=\frac{1}{2^4}=\frac{1}{16}\)

 1/2 x 2 mũ n cộng 4 x 2 mũ n = 9 x 2 mũ n