Đáp án:

 

Giải thích các bước giải:

$x²-4xy+5y²+6x-16y+12=0$

$→[(x²-4xy+4y²)+(6x-12y)+9]+(y²-4y+4)-1=0$

$→ [(x²-4xy+4y²)+(6x-12y)+9]+(y²-4y+4)=1$

$→ [(x-2y)²+6(x-2y)+9]+(y-2)²=1$

$→ (x-2y+3)²+(y-2)²=1$

$TH1:$

$(y-2)²=0$

$→y=2$

$(x-2y+3)²=1$

$→x-2y+3=±1$

$→x-2.2+3=±1$

$→x-1=±1$

$→x=0$ hoặc $x=2$

$TH2:$

$(y-2)²=1$

$→y-2=±1$

$→y=3$ hoặc $y=1$

Với $y=3:$

$(x-2y+3)²=0$

$→x-2y+3=0$

$→x-2.3+3=0$

$→x-3=0$

$→x=3$

Với $y=1$

$(x-2y+3)²=0$

$→x-2y+3=0$

$→x-2.1+3=0$

$→x+1=0$

$→x=-1$

Vậy....

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$x²-4xy+3y²+2x-8y-4=0$

$→[(x²-4xy+4y²)+(2x-4y)+1]+(-y²-4y-4)-1=0$

$→[(x²-4xy+4y²)+(2x-4y)+1]-(y²+4y+4)=1$

$→[(x-2y)²+2(x-2y)+1]-(y+2)²=1$

$→(x-2y+1)²-(y+2)²=1$

$→(x-y+3)(x-3y-1)=1$

$TH1:$

$x-y+3=1$

$→x-y=-2$

$x-3y-1=1$

$→x-y-2y=2$

$→-2-2y=2$

$→-2y=4$

$→y=-2$

$→x=-4$

$TH2:$

$x-y+3=-1$

$→x-y=2$

$x-3y-1=-1$

$→x-y-2y=0$

$→2-2y=0$

$→2y=2$

$→y=1$

$→x=3$

Vậy......