`a)`

$H(x) = x^4 + 2x^3 - x^4 + 2x^2 - x^3 - 3x - 6$

$H(x) = (x^4 - x^4) + (2x^3 - x^3) + 2x^2 - 3x - 6$

$H(x) = x^3 + 2x^2 - 3x - 6$

Bậc `3`

Hệ số cao nhất `1`

Hệ số tự do `-6`

`b)` 

$A(x) = -x^3 + 3x^2 - x + 5$

$B(x) = x^3 - 2x^2 + 3x - 4$

$M(x) = A(x) + B(x)$

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

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

$M(x) = x^2 + 2x + 1$

`c)`

Thay `x = 1/2` vào `C` ta có: 

`C(\frac{1}{2}) = 3(\frac{1}{2})^2 - \frac{1}{2} + 2`

`C(\frac{1}{2}) = 3 . \frac{1}{4} - \frac{1}{2} + 2`

`C(\frac{1}{2}) = \frac{3}{4} - \frac{1}{2} + 2`

`C(\frac{1}{2}) = \frac{3}{4} - \frac{2}{4} + \frac{8}{4}`

`C(\frac{1}{2}) = \frac{9}{4}`