tìm x,y 1,$x^{2}$ +$y^{2}$+6x-10y+34=0 2, $x^{2}$ +9$y^{2}$ +20x-6y+26=0 3,9$x^{2}$ +12x+4$y^{2}$+8y+8=0 4,4 $x^{2}$ +9$y^{2}$ +20-6y+26=0 5,3$x^{2}$

0 bình luận về “tìm x,y 1,$x^{2}$ +$y^{2}$+6x-10y+34=0 2, $x^{2}$ +9$y^{2}$ +20x-6y+26=0 3,9$x^{2}$ +12x+4$y^{2}$+8y+8=0 4,4 $x^{2}$ +9$y^{2}$ +20-6y+26=0 5,3$x^{2}$”

1. 1,  x²+y²+6x-10y+34=0

   ⇔ (x²+2.3x+9) + (y²-2.5y+25)=0 

   ⇔ (x+3)²+ (y-5)²=0 

   ⇔ (x+3)²=0 hoặc (y−5)²=0

   ⇔ x+3=0 hoặc y−5=0

   ⇔ x=−3 hoặc y =5

   3, 9x²+12x+4y²+8y+8=0

   ⇔ (9x²+2.3.2x+4)+(4y²+2.2.2y+4)=0 

   ⇔ (3x+2)² + (2y+2)² = 0

   ⇔ (3x+2²)=0 hoặc (2y+2)²=0

   ⇔ 3x+2=0 hoặc 2y+2=0

   ⇔ x=$\frac{-2}{3}$ hoặc y=−1

   4,   4x²+9y²+20x-6y+26=0

   ⇔ (4x²+2.2.5x+25)+(9y²-2.3y+1)=0

   ⇔ (2x+5)²+(3y-1)²=0

   ⇔ 2x+5=0 hoặc3y−1=0

   ⇔ x=$\frac{-5}{2}$ hoặc y=$\frac{1}{3}$

   5, 3x²+3y²+6x-12y+15=0

   ⇔ 3(x²+2x+1)+3(y²-4y+4)=0

   ⇔ 3(x+1)²+3(y-2)²=0

   ⇔ x+1=0 hoặc y−2=0

   ⇔ x=−1 hoặc y=2

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2. 1)    x²+y²+6x-10y+34=0

   <=>(x²+2.3x+9)+(y²-2.5y+25)=0

   <=>(x+3)²+(y-5)²=0

   <=>$\left \{ {{(x+3)²=0} \atop {(y-5)²=0}} \right.$ 

   <=>$\left \{ {{x+3=0} \atop {y-5=0}} \right.$ 

   <=>$\left \{ {{x=-3} \atop {y=5}} \right.$ 

   3)    9x²+12x+4y²+8y+8=0

   <=>(9x²+2.3.2x+4)+(4y²+2.2.2y+4)=0

   <=>(3x+2)²+(2y+2)²=0

   <=>$\left \{ {{(3x+2²)=0} \atop {(2y+2)²=0}} \right.$ 

   <=>$\left \{ {{3x+2=0} \atop {2y+2=0}} \right.$ 

   <=>$\left \{ {{x=-2/3} \atop {y=-1}} \right.$ 

   4)   4x²+9y²+20x-6y+26=0

   <=>(4x²+2.2.5x+25)+(9y²-2.3y+1)=0

   <=>(2x+5)²+(3y-1)²=0

   <=>$\left \{ {{2x+5=0} \atop {3y-1=0}} \right.$ 

   <=>$\left \{ {{x=-5/2} \atop {y=1/3}} \right.$ 

   5)     3x²+3y²+6x-12y+15=0

   <=>3(x²+2x+1)+3(y²-4y+4)=0

   <=>3(x+1)²+3(y-2)²=0

   <=>$\left \{ {{x+1=0} \atop {y-2=0}} \right.$ 

   <=>$\left \{ {{x=-1} \atop {y=2}} \right.$ 

   6)     x²+4y²+4x-4y+5=0

   <=>(x²+4x+4)+(4y²-4y+1)=0

   <=>(x+2)²+(2y-1)²=0

   <=>$\left \{ {{x+2=0} \atop {2y-1=0}} \right.$ 

   <=>$\left \{ {{x=-2} \atop {y=1/2}} \right.$ 

    

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