Question

[~ Kammin Meau ~]

1. Tính:

D = 3/5.7 + 3/7.9 + 3/9.11 + ... + 3/53.55

2. Tìm x sao cho:

A = x+2/x-5 là số nguyên dương

3. Chứng tỏ rằng:

S = 1/2 + 1/2^2 + 1/2^3 + ... + 1/2^20    nhỏ hơn 1

A = 1/10 + 1/11 + 1/12 + ... + 1/99 + 1/100    lớn hơn 1

 Giúp mình với. Hu hu

Answer 1

[Nguyễn Lê Phước Thịnh]

1: 

Ta có: \(D=\dfrac{3}{5\cdot7}+\dfrac{3}{7\cdot9}+\dfrac{3}{9\cdot11}+...+\dfrac{3}{53\cdot55}\)

\(=\dfrac{3}{2}\left(\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}+...+\dfrac{2}{53\cdot55}\right)\)

\(=\dfrac{3}{2}\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+...+\dfrac{1}{53}-\dfrac{1}{55}\right)\)

\(=\dfrac{3}{2}\left(\dfrac{1}{5}-\dfrac{1}{55}\right)\)

\(=\dfrac{3}{2}\left(\dfrac{11}{55}-\dfrac{1}{55}\right)\)

\(=\dfrac{3}{2}\cdot\dfrac{2}{11}=\dfrac{3}{11}\)

Answer 2

[Nguyễn Lê Phước Thịnh]

2) Để A là số nguyên dương thì 

\(\left\{{}\begin{matrix}x+2⋮x-5\\x-5>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-5+7⋮x-5\\x>5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}7⋮x-5\\x>5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-5\inƯ\left(7\right)\\x>5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-5\in\left\{1;-1;7;-7\right\}\\x>5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\in\left\{6;4;12;-2\right\}\\x>5\end{matrix}\right.\)

\(\Leftrightarrow x\in\left\{6;12\right\}\)