x(x-1)2-x2(x-1)+10
=x(x2-2x+1)-x3+x2+10
=x3-2x2+x-x3+x2+10
=-x2+x+10
=-(x2-x)+10.
∵x2-x-6=0,
∴x2-x=6,
∴原式=4.
已知x2-x-6=0,求代数式x(x-1)2-x2(x-1)+10的值.
已知x2-x-6=0,求代数式x(x-1)2-x2(x-1)+10的值.
数学人气:571 ℃时间:2019-08-22 14:30:18
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