Question

Ét o ét

Answer 1

a)\(A\left(x\right)=4x^4-x^2-x+1\)

\(B\left(x\right)=-3x^4-6x^3+x^2-\dfrac{2}{3}x-\dfrac{1}{2}\)

 

Answer 2

b)\(C\left(x\right)=4x^4-x^2-x+1-3x^4-6x^3+x^2-\dfrac{2}{3}x-\dfrac{1}{2}\)

\(C\left(x\right)=x^4-6x^3-\dfrac{5}{3}x+\dfrac{1}{2}\)

\(D\left(x\right)=4x^4-x^2-x+1+3x^4+6x^3-x^2+\dfrac{2}{3}x+\dfrac{1}{2}\)

\(D\left(x\right)=7x^4+6x^3-2x^2-\dfrac{1}{3}x+\dfrac{3}{2}\)

Answer 3

thay x= -1 vào C(x) ta đc

\(C\left(-1\right)=\left(-1\right)^4+6.\left(-1\right)^3-\dfrac{5}{3}.\left(-1\right)+\dfrac{1}{2}\)

\(C\left(-1\right)=1-6+\dfrac{5}{3}+\dfrac{1}{2}=-5+\dfrac{5}{3}+\dfrac{1}{2}=-\dfrac{17}{6}\)

mà \(-\dfrac{17}{6}\ne0\)

=> x = -1 ko là nghiệm của C(x)