Ta có: $(x-2)^2$+$(y-2)^2$-2xy\leq8
\Leftrightarrow$x^2$-4x+$y^2$+4y-2xy\leq0 (1)
\Leftrightarrow$(x-y-2)^2$\leq4\Leftrightarrow0<x-y\leq4
A=(x-y)($x^2$+xy+$y^2$)-(x-y)(7+3xy)+$\dfrac{4}{x-y}$+4xy
0<x-y\leq4\Rightarrow$\dfrac{4}{x-y}$\geq1
\RightarrowA\geq(x-y)[$(x-y)^2$-7]+4xy+1
Mặt khác 4\geqx-y\Leftrightarrow4(x-y)\geq$(x-y)^2$ (vì x-y>0)
\Leftrightarrowx-y\geq$\dfrac{(x-y)^2}{4}$
Vì x-7>$\sqrt{7}$\Rightarrow$(x-y)^2$-7>0 (vì 2 vế dương)
\RightarrowA\geq$\dfrac{(x-y)^2}{4}$[$(x-y)^2$-7]+4xy+1
\LeftrightarrowA\geq$\dfrac{(x-y)^4}{4}$-$\dfrac{7}{4}(x-y)^2$+4xy+1
Xét T= $(a-b)^4$ (ab>0)
=$a^4$+$4a^3b$+$6a^2b^2$+$4ab^3$+$b^4$
T\geq$2a^2b^2$+$6a^2b^2$+4ab($a^2+b^2$)
T\geq$2a^2b^2$+$6a^2b^2$+$8a^2b^2$ (vì ab>0)
\LeftrightarrowT\geq$16a^2b^2$
xy<0\Rightarrow-xy>0\Rightarrow$\dfrac{(x-y)^4}{4}$\geq$4x^2y^2$
\RightarrowA\geq$4x^2y^2$+4xy+1-$\dfrac{7}{4}(x-y)^2$
\LeftrightarrowA\geq$(2xy+1)^2$-$\dfrac{7}{4}(x-y)^2$
x-y\leq4\RightarrowA\geq$(2xy+1)^2$-28
Mặt khác (1)\Leftrightarrow2xy\geq$x^2$-4x+$y^2$-4y
\RightarrowA\geq$(x^2-4x+y^2+4y+1)^2$-28
\LeftrightarrowA\geq$[(x-2)^2+(y+2)^2-7]^2$-28\geq49-28=21
Vậy $A_{min}$=21\Leftrightarrowx=2 ; y=-2
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