Bài 5: fta có : f(100)-> x=100 -> x+1=101
-> F(100)=${x^8-(x+1)x^7+(x+1)x^6-(x+1)x^5+...+(x+1)x^2-(x+1)x+25=x^8-x^8+x^7-x^7+...+x^3+X^2-x^2+x+25=x+25=125}$
Bài 6 : f(1)=g(2) <=>${2.1^2+a+4=2^2-5.2-b<=>2+a+4-4+10+b=0 <=>a+b=-12}$
f(-1)=g(5)<=>${2-a+4=25-25-b<=>b-a=6}$
Có a+b=12; a-b=6 -> a=3, b=9