a)\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{99.101}=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}=1-\dfrac{1}{101}=\dfrac{100}{101}\)
b)ĐK: \(n\ne-5\)
\(A=\dfrac{n-2}{n+5}=\dfrac{n+5-7}{n+5}=1-\dfrac{7}{n+5}\)
Để A nguyên thì \(\dfrac{n-2}{n+5}\)phải nguyên <=> \(\dfrac{7}{n+5}\) nguyên mà n là số nguyên <=> 7 chia hết cho n+5 hay n+5 là Ư(7)
Mà Ư(7)={-1;1;-7;7}
Ta có bảng sau:
| n+5 | -1 | 1 | -7 | 7 |
| n | -6(TM) | -4(TM) | -12(TM) | 2(TM) |
Vậy n={-6;-4;-12;2} thì A nguyên
a. \(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}\)
\(=\dfrac{99}{100}\)
b, Ta có: \(A=\dfrac{n-2}{n+5}=\dfrac{n+5-7}{n+5}=1-\dfrac{7}{n+5}\)
Để \(A\in Z\) thì \(\dfrac{n-2}{n+5}\in Z\Rightarrow7⋮n+5\Leftrightarrow n+5\in U\left(7\right)=\left\{\pm1;\pm7\right\}\)
Lập bảng giá trị:
| \(n+5\) | \(1\) | \(-1\) | \(7\) | \(-7\) |
| \(n\) | \(-4\) | \(-6\) | \(2\) | \(-12\) |
Vậy, với \(x\in\left\{-12;-6;-4;2\right\}\) thì \(A=\dfrac{n-2}{n+5}\in Z\)
a)=1-1/3+1/3-1/5+1/5-1/7+....+1/99-1/101
=100/101
b) Để A e Z
<=> n-2 chia hết n+5
=>n-2=(x+5)-7 chia hết n+5
=>n+5 e Ư(7)
=>n+5 e{7;-7;1;-1}
=>n e {2;-12;-4;-6}
