Question

Số nguyên dương x lớn nhất để phân thức \(A=\dfrac{x^3+x-2}{x^2-3x^2-2x-8}\) có giá trị là số nguyên

Answer 1

\(A=\dfrac{x^3+x-2}{x^3-3x^2-2x-8}\)

\(=\dfrac{x^3-x+2x-2}{x^2-4x^2+x^2-4x+2x-8}\)

\(=\dfrac{x\left(x^2-1\right)+2\left(x-1\right)}{x^2\left(x-4\right)+x\left(x-4\right)+2\left(x-4\right)}\)

\(=\dfrac{x\left(x-1\right)\left(x+1\right)+2\left(x-1\right)}{\left(x-4\right)\left(x^2+x+2\right)}\)

\(=\dfrac{\left(x-1\right)\left[x\left(x+1\right)+2\right]}{\left(x-4\right)\left[x\left(x+1\right)+2\right]}\)

\(=\dfrac{x-4+3}{x-4}\)

\(=1+\dfrac{3}{x-4}\)

\(A\in Z\Leftrightarrow3⋮x-4\Leftrightarrow x-4\in\text{Ư}\left(3\right)=\left\{-3;-1;1;3\right\}\Leftrightarrow x\in\left\{1;3;5;7\right\}\)

mà x lớn nhất

=> x = 7