Bài 5 :
a) \(P\left(0\right)=4\Leftrightarrow c=4\)
\(P\left(1\right)=8\Leftrightarrow a+b+c=8\Leftrightarrow a+b=8-c=8-4=4\left(1\right)\)
\(P\left(-1\right)=10\Leftrightarrow a-b+c=10\Leftrightarrow a-b=10-c=10-4=6\left(2\right)\)
\(\left(1\right);\left(2\right)\Rightarrow2a=10\Leftrightarrow a=5\)
\(\left(1\right)\Rightarrow b=4-a=4-5=-1\)
Vậy \(\left(a;b;c\right)=\left(5;-1;4\right)\)
b) Theo trên ta được \(P\left(x\right)=5x^2-x+4\)
\(P\left(2\right)=5.2^2-2+4=22\)
\(P\left(3\right)=5.3^2-3+4=46\)
\(K=\dfrac{P\left(2\right)-P\left(3\right)}{12}=\dfrac{22-46}{12}=-2\)
Bài 6:
x=1000 nên x+1=1001
\(P\left(x\right)=x^8-1001x^7+1001x^6-1001x^5+...+1001x^2-1001x+250\)
\(=x^8-x^7\left(x+1\right)+x^6\left(x+1\right)-x^5\left(x+1\right)+...+x^2\left(x+1\right)-x\left(x+1\right)+250\)
\(=x^8-x^8-x^7+x^7+...+x^3+x^2-x^2-x+250\)
=-x+250
=-1000+250
=-750