x²+x-6=0
(x-2)(x+3)=0
x1=2,x2=-3
当x=2时x²(x+1)-x(x²-1)-7=4*3-2*3-7=-1
当x=-3时x²(x+1)-x(x²-1)-7=9*(-2)+3*8-7=-1
x²(x+1)-x(x²-1)-7=x^3+x²-x^3+x-7=x²+x-7=(x²+x-6)-1=-1
解:x2(x+1)-x(x2-1)-7,
=x3+x2-x3+x-7,
=x2+x-7,
∵x2+x-6=0,
∴x2+x-7=-1,
即x2(x+1)-x(x2-1)-7=-1.
x^2(x+1)-x(x^2-1)-7=x^3+x^2-x^3+x-7=(x^2+x-6)-1
因x^2+x-6=0,所以原式=0-1=-1
x²+x-6=0,即x(x+1)=6.
x²(x+1)-x(x²-1)-7=x*x(x+1)-(x-1)*x(x+1)-7=6x-6(x-1)-7=-1
x²+x=6
原式=x³+x²-x³+x-7
=x²+x-7
=6-7
=-1