Giải:
Đặt: \(A=\dfrac{x^3+2x^2+5x+10}{x^2+4x+4}\)
\(\Leftrightarrow A=\dfrac{x^2\left(x+2\right)+5\left(x+2\right)}{\left(x+2\right)^2}\)
\(\Leftrightarrow A=\dfrac{\left(x+2\right)\left(x^2+5\right)}{\left(x+2\right)^2}\)
\(\Leftrightarrow A=\dfrac{x^2+5}{x+2}\)
Để \(A\in Z\) thì:
\(\dfrac{x^2+5}{x+2}\in Z\)
\(\Leftrightarrow x^2+5⋮x+2\)
\(\Leftrightarrow x^2+4x+4-4x+1⋮x+2\)
\(\Leftrightarrow1-4x⋮x+2\)
\(\Leftrightarrow9-4x-8⋮x+2\)
\(\Leftrightarrow9-\left(4x+8\right)⋮x+2\)
\(\Leftrightarrow9-4\left(x+2\right)⋮x+2\)
\(\Leftrightarrow9⋮x+2\)
\(\Leftrightarrow x+2\inƯ\left(9\right)=\left\{\pm1;\pm3;\pm9\right\}\)
\(\Leftrightarrow x=\left\{-3;-1;-5;1;-11;7\right\}\)
Vậy ...
Đặt A = \(\dfrac{x^3+2x^2+5x+10}{x^2+4x+4}\) (x khác -2)
\(=\dfrac{x\left(x+2\right)+5\left(x+2\right)}{\left(x+2\right)^2}\)
\(=\dfrac{\left(x+2\right)\left(x+5\right)}{\left(x+2\right)^2}\)
\(=\dfrac{x+5}{x+2}=1+\dfrac{3}{x+2}\)
Để A nguyên thì 3 chia hết cho x+2 \(\Rightarrow x+2\in\left\{-3;-1;1;3\right\}\Rightarrow x\in\left\{-5;-3;-1;1\right\}\)