Question

[Soái muội]

Tìm số hữ tỉ a,b sao cho:

\(x^3+ax+b\)chia cho \(x+1\)dư 7,chia cho \(x-3\)dư 5

Answer 1

[zZz Cool Kid_new zZz]

Ta có:

\(x^3+ax+b=\left(x+1\right)\cdot P\left(x\right)+7\)

\(x^3+ax+b=\left(x-3\right)\cdot Q\left(x\right)+5\)

Theo Bezut ta có:

Với \(x=-1\Rightarrow b-a-1=7\)

Với \(x=3\Rightarrow3a+b+27=5\)

\(\Rightarrow4a+28=-2\Rightarrow4a=26\Rightarrow a=\frac{13}{2}\Rightarrow b=\frac{29}{2}\)