求方程5x^2+5y^2+8xy+2y-2x+2=0的实数解
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求方程5x^2+5y^2+8xy+2y-2x+2=0的实数解求方程5x^2+5y^2+8xy+2y-2x+2=0的实数解求方程5x^2+5y^2+8xy+2y-2x+2=0的实数解5x²+5
求方程5x^2+5y^2+8xy+2y-2x+2=0的实数解
求方程5x^2+5y^2+8xy+2y-2x+2=0的实数解
求方程5x^2+5y^2+8xy+2y-2x+2=0的实数解
5x²+5y²+8xy+2y-2x+2=0
4x²+4y²+8xy+y²+2y+1+x²-2x+1=0
(x-1)²+(y+1)²+(2x+2y)²=0
所以可得:
x-1=0
y+1=0
2x+2y=0
即实数
x=1
y=-1
原式因式分解,可得4*(x+y)^2+(x-1)^2+(y+1)^2=0;
嗯,化到这里,应该不用再继续了吧。像这种明显条件不够的,一定是因式分解有特殊性的
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