a)\(A\left(x\right)=4x^4-x^2-x+1\)
\(B\left(x\right)=-3x^4-6x^3+x^2-\dfrac{2}{3}x-\dfrac{1}{2}\)
b)\(C\left(x\right)=4x^4-x^2-x+1-3x^4-6x^3+x^2-\dfrac{2}{3}x-\dfrac{1}{2}\)
\(C\left(x\right)=x^4-6x^3-\dfrac{5}{3}x+\dfrac{1}{2}\)
\(D\left(x\right)=4x^4-x^2-x+1+3x^4+6x^3-x^2+\dfrac{2}{3}x+\dfrac{1}{2}\)
\(D\left(x\right)=7x^4+6x^3-2x^2-\dfrac{1}{3}x+\dfrac{3}{2}\)
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)
