Question

Answer 1

\(a,A=\dfrac{3}{x^2-2x+2}=\dfrac{3}{\left(x-1\right)^2+1}\le\dfrac{3}{0+1}=3\\ A_{max}=3\Leftrightarrow x-1=0\Leftrightarrow x=1\\ b,A=\dfrac{3}{x^2-2x+2}\Leftrightarrow Ax^2-2Ax+2A-3=0\\ \Leftrightarrow\Delta'=A^2-A\left(2A-3\right)=-A\left(A-3\right)\ge0\\ \Leftrightarrow0\le A\le3\)

Do đó \(A\in Z\Leftrightarrow A\in\left\{0;1;2;3\right\}\)

Với \(A=0\Leftrightarrow3=0\left(\text{vô lí}\right)\)

Với \(A=1\Leftrightarrow x^2-2x+2=3\Leftrightarrow x^2-2x-1=0\Leftrightarrow x=1\pm\sqrt{2}\)

Với \(A=2\Leftrightarrow2x^2-4x+4=3\Leftrightarrow2x^2-4x+1=0\Leftrightarrow x=\dfrac{2\pm\sqrt{2}}{2}\)

Với \(A=3\Leftrightarrow x^2-2x+2=1\Leftrightarrow x^2-2x+1=0\Leftrightarrow x=1\)

Vậy \(A\in Z\Leftrightarrow x\in\left\{1\pm\sqrt{2};\dfrac{2\pm\sqrt{2}}{2};1\right\}\)