Question

[Đinh Chấn Thiên]

a)\(\dfrac{2}{1x3}+\dfrac{2}{3x5}+\dfrac{2}{5x7}...+\dfrac{2}{99x101}\)

b)Cho A=\(\dfrac{n-2}{n+5}\)(neZ;n#5) Tìm n để AeZ

Answer 1

[Xuân Tuấn Trịnh]

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

Answer 2

[Sáng]

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 \(A=\dfrac{n-2}{n+5}\in Z\)