Question

1) Cho P(x)= 100x^100 + 99x^99 +...+ 2x^2 + x

Tính P(1);P(-1)

2)Cho Q (x) = x^99 - 100x^99 + 100x^97 + 100x^96

Tính Q(99)

3)Cho 2 đa thức:

P(x)=x^2 + 2 nx + m^2

Q(x)=x^2 + (2m+1) x + m^2

Tìm m biết P(1)=Q(-1)

4)Ch P(x) = ax^2 + bx + c

Chứng tỏ P(-1).P(-2) < hoặc =0

Answer 1

Câu 2: 

Sửa đề; \(Q\left(x\right)=x^{99}-100x^{98}+100x^{97}-100x^{96}\)

x=99 nên x+1=100

\(Q\left(x\right)=x^{99}-x^{98}\left(x+1\right)+x^{97}\left(x+1\right)-x^{96}\left(x+1\right)\)

\(=x^{99}-x^{99}-x^{98}+x^{98}+x^{97}-x^{97}-x^{96}\)

\(=-x^{96}=-99^{96}\)