Bài 11:
d: \(A=\dfrac{1}{1\cdot4}+\dfrac{1}{4\cdot7}+...+\dfrac{1}{22\cdot25}\)
\(=\dfrac{1}{3}\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{22\cdot25}\right)\)
\(=\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{22}-\dfrac{1}{25}\right)\)
\(=\dfrac{1}{3}\left(1-\dfrac{1}{25}\right)=\dfrac{1}{3}\cdot\dfrac{24}{25}=\dfrac{8}{25}\)
\(B=\left(1+\dfrac{1}{1\cdot3}\right)\left(1+\dfrac{1}{2\cdot4}\right)\cdot...\cdot\left(1+\dfrac{1}{22\cdot24}\right)\left(1+\dfrac{1}{23\cdot25}\right)\)
\(=\left(1+\dfrac{1}{2^2-1}\right)\left(1+\dfrac{1}{3^2-1}\right)\cdot...\cdot\left(1+\dfrac{1}{24^2-1}\right)\)
\(=\dfrac{2^2-1+1}{2^2-1}\cdot\dfrac{3^2-1+1}{3^2-1}\cdot...\cdot\dfrac{24^2-1+1}{24^2-1}\)
\(=\dfrac{2^2\cdot3^2\cdot...\cdot24^2}{\left(2^2-1\right)\left(3^2-1\right)\cdot...\cdot\left(24^2-1\right)}\)
\(=\dfrac{2\cdot3\cdot...\cdot24}{1\cdot2\cdot...\cdot23}\cdot\dfrac{2\cdot3\cdot...\cdot24}{3\cdot4\cdot...\cdot25}=\dfrac{24}{1}\cdot\dfrac{2}{25}=\dfrac{48}{25}\)
\(3A< \dfrac{8\left(x-3\right)}{25}< B\)
=>\(3\cdot\dfrac{8}{25}< \dfrac{8\left(x-3\right)}{25}< \dfrac{48}{25}\)
=>24<8(x-3)<48
=>3<x-3<6
=>6<x<9
mà x nguyên
nên \(x\in\left\{7;8\right\}\)