作业帮姹傛柟绋求方程5x2-6xy+2y2-4x+2y+1=0的实数根
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姹傛柟绋求方程5x2-6xy+2y2-4x+2y+1=0的实数根
姹傛柟绋求方程5x2-6xy+2y2-4x+2y+1=0的实数根
姹傛柟绋求方程5x2-6xy+2y2-4x+2y+1=0的实数根
方法一:
5x²-6xy+2y²-4x+2y+1
=(4x²-4xy+y²)+(-4x+2y)+1+x²+y²-2xy
=(2x-y)²-2(2x-y)+1+(x-y)²
=(2x-y-1)²+(x-y)²
所以,2x-y-1=0且x=y,
解得,x=1,y=1
方法二:
5x²-6xy+2y²-4x+2y+1=0
5x²+(-6y-4)x+(2y²+2y+1)=0
△=(-6y-4)²-4×5×(2y²+2y+1)=-4y²+8y-4
令-4y²+8y-4=0
解得y=1
代入原方程,解得x=1
所以,方程5x²-6xy+2y²-4x+2y+1=0的实数根为x=1,y=1
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