已知a^2-3a+1= 0,则:(1)a^2+1/a^2= (2)a^2/(a^4+a^2+1已知a^2-3a+1= 0,则:(1)a^2+1/a^2= (2)a^2/(a^4+a^2+1)=

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已知a^2-3a+1=0,则:(1)a^2+1/a^2=(2)a^2/(a^4+a^2+1已知a^2-3a+1=0,则:(1)a^2+1/a^2=(2)a^2/(a^4+a^2+1)=已知a^2-3a

已知a^2-3a+1= 0,则:(1)a^2+1/a^2= (2)a^2/(a^4+a^2+1已知a^2-3a+1= 0,则:(1)a^2+1/a^2= (2)a^2/(a^4+a^2+1)=
已知a^2-3a+1= 0,则:(1)a^2+1/a^2= (2)a^2/(a^4+a^2+1
已知a^2-3a+1= 0,则:(1)a^2+1/a^2= (2)a^2/(a^4+a^2+1)=

已知a^2-3a+1= 0,则:(1)a^2+1/a^2= (2)a^2/(a^4+a^2+1已知a^2-3a+1= 0,则:(1)a^2+1/a^2= (2)a^2/(a^4+a^2+1)=
已知a^2−3a+1=0,则:a^2/(a^4+a^2+1)=?a^2−3a+1=0,有 a^2+a+1=4a; a^2−a+1=2a.a^2/(a^4+a^2+1)=a^2/(a^4+2a^2+1−a^2)=a^2/[(a^2+1)^2−a^2]=a^2/[(a^2+a+1)(a^2−a+1)]=a^2/(4a·2a)=1/8