Question

[Lê Hoàng ThủyTiên]

 giúp mình với ! chiều ni thầy kiểm tra rồi SOS

Answer 1

[HT.Phong (9A5)]

Bài 3:

\(A=75\left(4^{2004}+4^{2003}+...+4^2+4+1\right)+25\)

Đặt: \(B=4^{2004}+4^{2003}+...+4^2+4+1\)

\(4B=4^{2005}+4^{2004}+...+4^3+4^2+4\\ 4B-B=\left(4^{2005}+4^{2004}+...+4^3+4^2+4\right)-\left(4^{2004}+4^{2003}+...+4^2+4+1\right)\\ 3B=4^{2005}-1\\ B=\dfrac{4^{2005}-1}{3}\)

\(=>A=75\cdot\dfrac{4^{2005}-1}{3}+25\\ =25\left(4^{2005}-1\right)+25\\ =25\cdot\left(4^{2005}-1+1\right)\\ =25\cdot4^{2005}\\ =25\cdot4\cdot4^{2004}\\ =100\cdot4^{2004}\) 

=> A chia hết cho 100 

Answer 2

[Nguyễn Lê Phước Thịnh]

Bài 1:

a: \(\dfrac{5\cdot3^{11}+4\cdot3^{17}}{3^9\cdot5^2-3^9\cdot2^3}=\dfrac{3^{11}\cdot\left(5+4\cdot3^6\right)}{3^9\left(5^2-2^3\right)}\)

\(=3^2\cdot\dfrac{5+4\cdot729}{25-8}=3^2\cdot\dfrac{2921}{17}=\dfrac{26289}{17}\)

b: \(\dfrac{1}{2\cdot5}+\dfrac{1}{5\cdot8}+...+\dfrac{1}{47\cdot50}\)

\(=\dfrac{1}{3}\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+...+\dfrac{3}{47\cdot50}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{47}-\dfrac{1}{50}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{50}\right)=\dfrac{1}{3}\cdot\dfrac{24}{50}=\dfrac{24}{150}=\dfrac{8}{50}=\dfrac{4}{25}\)

c: \(1+\dfrac{2}{6}+\dfrac{2}{12}+...+\dfrac{2}{9900}\)

\(=\dfrac{2}{2}+\dfrac{2}{6}+...+\dfrac{2}{9900}\)

\(=2\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{9900}\right)\)

\(=2\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\right)\)

\(=2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(=2\left(1-\dfrac{1}{100}\right)=2\cdot\dfrac{99}{100}=\dfrac{99}{50}\)

d: \(\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)\cdot...\cdot\left(\dfrac{1}{99^2}-1\right)\)

\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{99}-1\right)\left(\dfrac{1}{2}+1\right)\cdot\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{99}+1\right)\)

\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-98}{99}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{100}{99}\)

\(=\dfrac{1}{99}\cdot\dfrac{100}{2}=\dfrac{50}{99}\)