Đáp án:
$a) A(x)=3x^4-2x^3+x^2-\frac{5}{2}\\
B(x)=-3x^4+2x^3-x^2-x+\frac{5}{2}\\
b)
A(x)+B(x)=-x\\
A(x)+B(x)=6x^4-4x^3+2x^2+x-5\\
c) x=0$
Giải thích các bước giải:
$a) A(x)=\frac{-5}{2}+3x^4-2x^3+x^2\\
=3x^4-2x^3+x^2-\frac{5}{2}\\
B(x)=2x^3+\frac{5}{2}-3x^4-x-x^2\\
=-3x^4+2x^3-x^2-x+\frac{5}{2}\\
b)
A(x)+B(x)=3x^4-2x^3+x^2-\frac{5}{2}+(-3x^4+2x^3-x^2-x+\frac{5}{2})\\
=3x^4-2x^3+x^2-\frac{5}{2}-3x^4+2x^3-x^2-x+\frac{5}{2}\\
=(3x^4-3x^4)+(-2x^3+2x^3)+(x^2-x^2) -x +(\frac{-5}{2}+\frac{5}{2})\\
=-x\\
A(x)+B(x)=3x^4-2x^3+x^2-\frac{5}{2}-(-3x^4+2x^3-x^2-x+\frac{5}{2})\\
=3x^4-2x^3+x^2-\frac{5}{2}+3x^4-2x^3+x^2+x-\frac{5}{2}\\
=(3x^4+3x^4)+(-2x^3-2x^3)+(x^2+x^2) +x +(\frac{-5}{2}-\frac{5}{2})\\
=6x^4-4x^3+2x^2+x-5\\
c)
A(x)+B(x)=0\\
\Leftrightarrow -x=0\\
\Leftrightarrow x=0$
