Question

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

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

2) Để A là số nguyên dương thì 

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

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

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

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

Answer 3

Giải:

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

   D=3/2.(2/5.7+2/7.9+2/9.11+...+2/53.55)

   D=3/2.(1/5-1/7+1/7-1/9+1/9-1/11+...+1/53-1/55)

   D=3/2.(1/5-1/55)

   D=3/2.2/11

   D=3/11

Answer 4

2) A=x+2/x-5 là số nguyên dương thì x+2 ⋮ x-5 

x+2 ⋮ x-5

x-5+7 ⋮ x-5

⇒7 ⋮ x-5

⇒x-5 ∈ Ư(7)={-7;-1;1;7}

Ta có bảng:

x-5=-7

   x=-2 (loại)

x-5=-1

   x=4 (t/m)

x-5=1

  x=6 (t/m)

x-5=7

   x=12 (t/m)

Chúc bạn học tốt!

Answer 5

3) S=1/2+1/2²+1/2³+...+1/2²⁰

  2S=1+1/2+1/2²+...+1/2¹⁹

2S-S=(1+1/2+1/2²+...+1/2¹⁹)-(1/2+1/2²+1/2³+...+1/2²⁰)

   S=1-1/2²⁰

Vì có ''1-'' tức S<1 nên S<1

Chúc bạn học tốt!