Có $-x^2+2x-2=-x^2+2x-1-1=-(x-1)^2-1$
$⇒A=\dfrac{1}{-x^2+2x-2}=\dfrac{1}{-(x-1)^2-1}=\dfrac{-1}{(x-1)^2+1}$
Có: $(x-1)^2≥0∀x⇒(x-1)^2+1≥1∀x⇒\dfrac{1}{(x-1)^2+1}≤\dfrac{1}{1}=1$
$⇒A=\dfrac{-1}{(x-1)^2+1}≥-1$
Dấu $=$ xảy ra $⇔(x-1)^2=0⇔x-1=0⇔x=1$
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