将原方程展开,得:x^2-ax-bx+ax-a^2-ab-1=0
进一步化简,得:x^2-bx-a^2-ab-1=0
判别式△=b^2+4(a^2+a+1)=b^2+4[(a+1/2)^2 + 3/4]
由于b^2 >= 0;(a+1/2)^2 + 3/4 >= 0;因此△ > 0.
故:原方程有两个不相等的实根.
判断关于x的一元二次方程(X+a)(x-a-b)=1根的情况,
判断关于x的一元二次方程(X+a)(x-a-b)=1根的情况,
数学人气:482 ℃时间:2019-11-01 18:49:56
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