若多项式x⁴-(a-1)x³+3x²-(b+1)x-2不含x³和x项,则a= b=
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![若多项式x⁴-(a-1)x³+3x²-(b+1)x-2不含x³和x项,则a= b=](/uploads/image/z/2590288-16-8.jpg?t=%E8%8B%A5%E5%A4%9A%E9%A1%B9%E5%BC%8Fx%26%238308%3B-%28a-1%29x%26%23179%3B%2B3x%26%23178%3B-%28b%2B1%29x-2%E4%B8%8D%E5%90%ABx%26%23179%3B%E5%92%8Cx%E9%A1%B9%2C%E5%88%99a%3D+b%3D)
若多项式x⁴-(a-1)x³+3x²-(b+1)x-2不含x³和x项,则a= b=
若多项式x⁴-(a-1)x³+3x²-(b+1)x-2不含x³和x项,则a= b=
若多项式x⁴-(a-1)x³+3x²-(b+1)x-2不含x³和x项,则a= b=
因为x⁴-(a-1)x³+3x²-(b+1)x-2不含x³和x项 所以x³和x的系数为0 所以a-1=0 a=1
b+1=0 b=-1
不含x³和x项则他们的系数为0
所以-(a-1)=
-(b+1)=0
所以a=1,b=-1