题目内容
用因式分解法解方程:(y-1)2+2y(1-y)=0.
解:∵(y-1)2+2y(1-y)=0,
∴(y-1)2-2y(y-1)=0.∴(y-1)(y-1-2y)=0.
∴y-1=0或y-1-2y=0.∴y1=1,y2=-1.
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题目内容
用因式分解法解方程:(y-1)2+2y(1-y)=0.
解:∵(y-1)2+2y(1-y)=0,
∴(y-1)2-2y(y-1)=0.∴(y-1)(y-1-2y)=0.
∴y-1=0或y-1-2y=0.∴y1=1,y2=-1.