Question

Tìm \(n\in Z\) để A = \(1:\left(\dfrac{1}{2011}-\dfrac{1}{2011+n}\right)\) có giá trị nguyên

Answer 1

Ta có:\(A=1:\left(\dfrac{1}{2011}-\dfrac{1}{2011+n}\right)\left(đk:n\ne-2011\right)\)

\(A=1:\dfrac{n}{2011\cdot\left(2011+n\right)}\)

\(A=1\cdot\dfrac{2011\cdot\left(2011+n\right)}{n}\)

\(A=\dfrac{2011\cdot\left(2011+n\right)}{n}\)

\(\Rightarrow A\in Z\)

\(\Leftrightarrow\dfrac{2011\cdot\left(2011+n\right)}{n}\in Z\)

\(\Leftrightarrow2011\cdot\left(2011+n\right)⋮n\)

\(\Leftrightarrow2011^2+2011n⋮n\)

\(\Leftrightarrow2011^2⋮n\)

\(\Leftrightarrow4044121⋮n\)

\(\Leftrightarrow n\inƯ\left(4044121\right)\)

\(\Leftrightarrow n\in\left\{\pm1;2011;\pm4044121\right\}\)

Answer 2

A = 1 : (1/2011 - 1/2011 - 1/n)

A = 1 : (0 - 1/n)

A = 1 : (-1/n)

Để A có gía trị nguyên thì -1/n phải là Ước của 1

=> -1/n = {-1;1}

=> n = 1;-1