Đáp án:
Giải thích các bước giải:
$a)x^2+y^2-2x+4y+5=0$
$⇔(x^2-2x+1)+(y^2+4y+4)=0$
$⇔(x-1)^2+(y+2)^2=0$
$⇔\left \{ {{y+2=0} \atop {x-1=0}} \right.$
$⇔\left \{ {{y=-2} \atop {x=1}} \right.$
$b)5x^2+9y^2-12xy-6x+9=0$
$⇔(4x^2-12xy+9y^2)+(x^2-6x+9)=0$
$⇔(2x-3y)^2+(x-3)^2=0$
$⇔\left \{ {{2x-3y=0} \atop {x-3=0}} \right.$
$⇔\left \{ {{2.3-3y=0} \atop {x=3}} \right.$
$⇔\left \{ {{y=2} \atop {x=3}} \right.$
$c)5x^3-3x^2+10x-6=0$
$⇔(5x^3-3x^2)+(10x-6)=0$
$⇔x^2.(5x-3)+2.(5x-3)=0$
$⇔(5x-3).(x^2+2)=0$
$⇔5x-3=0$ $( vì x^2+2>0)$
$⇔5x=3$
$⇔x=\dfrac{3}{5}$
$d)(x-2)^3-(x+2).(x^2-2x+4)+(2x-3).(3x-2)=0$
$⇔x^3-6x^2+12x-8-x^3-8+6x^2-13x+6=0$
$⇔-x-10=0$
$⇔-x=10$
$⇔x=-10$
