Giải thích các bước giải:

a.Ta có:

$P(x)=2x-4x^4-3x^3+5+x+2x^4+x^3+1-3x^2$ 

$\to P(x)=-4x^4+2x^4-3x^3+x^3-3x^2+2x+x+5+1$

$\to P(x)=-2x^4-2x^3-3x^2+3x+6$

Ta có:

$Q(x)=7x^4+2x^2+2x^3-7-5x^4+x^2-2x$

$\to Q(x)=7x^4-5x^4+2x^3+2x^2+x^2-2x-7$

$\to Q(x)=2x^4+2x^3+3x^2-2x-7$

Ta có:

$A(x)=P(x)+Q(x)$

$\to A(x)=(-2x^4-2x^3-3x^2+3x+6)+(2x^4+2x^3+3x^2-2x-7)$

$\to A(x)=-2x^4-2x^3-3x^2+3x+6+2x^4+2x^3+3x^2-2x-7$

$\to A(x)=-2x^4+2x^4-2x^3+2x^3-3x^2+3x^2+3x-2x+6-7$

$\to A(x)=x-1$

b.Ta có: $A(x)=0$

$\to x-1=0$

$\to x=1$

c.Ta có:

$B(x)=x(x-1)+1$

$\to B(x)=x^2-x+1$

$\to B(x)=x^2-\dfrac12x-\dfrac12x+\dfrac14+\dfrac34$

$\to B(x)=x(x-\dfrac12)-\dfrac12(x-\dfrac12)+\dfrac34$

$\to B(x)=(x-\dfrac12)^2+\dfrac34>0$

$\to B(x)$ vô nghiệm