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解方程组 2x²-xy+y²=2y;2x²+4xy=5y
人气:300 ℃ 时间:2020-02-10 10:26:34
解答
2x²-xy+y²=2y; 2x^2=2y+xy-y^2
2x²+4xy=5y 2x^2=5y-4xy
2y+xy-y^2=5y-4xy
5xy-3y-y^2=0
y*(5x-3-y)=0
(1)y=0,代入原式:x=0
(2)5x-3-y=0,y=5x-3,代入原式:
2x^2-x*(5x-3)+(5x-3)^2=2(5x-3)
2x^2=(5x-3)(2+x-5x+3)
22x^2-37x+15=0
(22x-15)(x-1)=0
x1=15/22,y1=5x1-3=9/22
x2=1,y2=5x2-3=2
所以:原方程解为三组:
(1)x1=0,y1=0
(2)x2=15/22,y2=9/22
(3)x3=1,y3=2
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