1:
Ta có: \(D=\dfrac{3}{5\cdot7}+\dfrac{3}{7\cdot9}+\dfrac{3}{9\cdot11}+...+\dfrac{3}{53\cdot55}\)
\(=\dfrac{3}{2}\left(\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}+...+\dfrac{2}{53\cdot55}\right)\)
\(=\dfrac{3}{2}\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+...+\dfrac{1}{53}-\dfrac{1}{55}\right)\)
\(=\dfrac{3}{2}\left(\dfrac{1}{5}-\dfrac{1}{55}\right)\)
\(=\dfrac{3}{2}\left(\dfrac{11}{55}-\dfrac{1}{55}\right)\)
\(=\dfrac{3}{2}\cdot\dfrac{2}{11}=\dfrac{3}{11}\)
2) Để A là số nguyên dương thì
\(\left\{{}\begin{matrix}x+2⋮x-5\\x-5>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-5+7⋮x-5\\x>5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}7⋮x-5\\x>5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-5\inƯ\left(7\right)\\x>5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-5\in\left\{1;-1;7;-7\right\}\\x>5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\in\left\{6;4;12;-2\right\}\\x>5\end{matrix}\right.\)
\(\Leftrightarrow x\in\left\{6;12\right\}\)
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
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!
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!