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$A=\dfrac{(x^3+y^3)-(x^2+y^2)}{(x-1)(y-1)}=\dfrac{(x^3-x^2)+(y^3-y^2)}{(x-1)(y-1)}=\dfrac{x^2}{y-1}+\dfrac{y^2}{x-1}$.
Ta biết rằng
$(x-2)^2 \ge 0\Rightarrow x^2 \ge 4(x-1)\Rightarrow \dfrac{x^2}{x-1} \ge 4.$
tương tự $\dfrac{y^2}{y-1} \ge 4.$
Vậy
$A=\dfrac{x^2}{y-1}+\dfrac{y^2}{x-1}\ge2\sqrt{\dfrac{x^2}{y-1}.\dfrac{y^2}{x-1}}=2\sqrt{\dfrac{x^2}{x-1}.\dfrac{y^2}{y-1}}\ge2\sqrt{4.4}=8$
Vậy $A \ge 8,$ đpcm.
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