a) \(A=\frac{2x^3+x^2+7x+1}{x^2+3}=\frac{2x\left(x^2+3\right)+x^2+3+x-2}{x^2+3}\)
\(=\frac{\left(x^2+3\right)\left(2x+1\right)+x-2}{x^2+3}=2x+1+\frac{x-2}{x^2+3}\)
Vì \(x\in Z\Rightarrow2x+1\in Z\) nên để \(A\) nguyên thì \(\frac{x-2}{x^2+3}\) nguyên
\(\Rightarrow\left(x-2\right)⋮\left(x^2+3\right)\)
\(\Rightarrow\left(x-2\right)\left(x+2\right)⋮\left(x^2+3\right)\)
\(\Leftrightarrow\left(x^2-4\right)⋮\left(x^2+3\right)\)
\(\Leftrightarrow\left(x^3+3-7\right)⋮\left(x^2+3\right)\)
\(\Rightarrow7⋮\left(x^2+3\right)\Leftrightarrow\left(x^2+3\right)\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
\(\Rightarrow x^2+3=7\Leftrightarrow x^2=4\Leftrightarrow x=\pm2\)
Thử lại thấy \(x=2\) thỏa mãn.
Vậy \(x=2\).
b) \(\frac{2x^3-3x^2+3x}{x^2+1}=\frac{\left(x^2+1\right)\left(2x-3\right)+x+3}{x^2+1}=2x-3+\frac{x+3}{x^2+1}\)
\(\Rightarrow\left(x+3\right)⋮\left(x^2+1\right)\)
\(\Rightarrow\left(x^2-9\right)⋮\left(x^2+1\right)\)
\(\Leftrightarrow\left(x^2+1-10\right)⋮\left(x^2+1\right)\)
\(\Rightarrow10⋮\left(x^2+1\right)\Leftrightarrow\left(x^2+1\right)\inƯ\left(10\right)=\left\{\pm1;\pm2;\pm5;\pm10\right\}\)
\(\Rightarrow x^2\in\left\{0;1;4;9\right\}\)
\(\Leftrightarrow x\in\left\{0;\pm1;\pm2;\pm3\right\}\)
Thử lại...
c) Hoàn toàn tương tự.
giúp mình với tối nay mình cần gấp ạ