因式分解法:(1+√2)x^2-(1-√2)x=0

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因式分解法:(1+√2)x^2-(1-√2)x=0

因式分解法:(1+√2)x^2-(1-√2)x=0
因式分解法:(1+√2)x^2-(1-√2)x=0

因式分解法:(1+√2)x^2-(1-√2)x=0

(1+√2)X^2-(1-√2)X=0
x[(1+√2)X-(1-√2)]=0
x=0,(1+√2)X-(1-√2)=0
于是解得
x=0
,或x=(1-√2)/(1+√2)=2√2-3
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因式分解?
怎么给的是一个等式呀!
楼主是要解方程吧!
如果是的话:


(1+√2)x^2-(1-√2)x=0
x[(1+√2)x-(1-√2)]=0
有:x=0、(1+√2)x-(1-√2)=0
解得:x1=0、x2=(1-√2)/(1+√2)=-(1-√2)^2=2√2-3