y=-x²+(2-p)x-1=-(1+px+x²)+2x
依题意α和β是方程y=0两个根,αβ=1,且
-(1+pα+α²)+2α=0; -(1+pβ+β²)+2β=0,
∴1+pα+α²=2α; 1+pβ+β²=2β,
那么(1+pα+α²)(1+pβ+β²)=2α*2β=4αβ=4.解毕.
附注:若是考试填空题,也可设p为某一数字如p=0,则y=-x²+2x-1=-(x-1)²,
得α=β=1,所求式=(1+1)(1+1)=4..
y=-x²+(2-p)x-1=-(1+px+x²)+2x
依题意α和β是方程y=0两个根,αβ=1,且
-(1+pα+α²)+2α=0; -(1+pβ+β²)+2β=0,
∴1+pα+α²=2α; 1+pβ+β²=2β,
那么(1+pα+α²)(1+pβ+β²)=2α*2β=4αβ=4.解毕.
附注:若是考试填空题,也可设p为某一数字如p=0,则y=-x²+2x-1=-(x-1)²,
得α=β=1,所求式=(1+1)(1+1)=4..