Đáp án:
c) \(Bậc:4\)
Giải thích các bước giải:
\(\begin{array}{l}
a)A + B = 4{x^2} - 5xy + 3{y^2} + 3{x^2} + 2xy - {y^2}\\
= 7{x^2} - 3xy + 2{y^2}\\
\to Bậc:2\\
A - B = 4{x^2} - 5xy + 3{y^2} - 3{x^2} - 2xy + {y^2}\\
= {x^2} - 7xy + 4{y^2}\\
\to Bậc:2\\
b)C + D = {x^3} - 2{x^2}y + \dfrac{1}{2}x{y^2} - {y^4} + 1 - {x^3} - \dfrac{1}{2}{x^2}y + x{y^2} - {y^4} - 2\\
= - \dfrac{5}{2}{x^2}y + \dfrac{3}{2}x{y^2} - 2{y^4} - 1\\
\to Bậc:4\\
C - D = {x^3} - 2{x^2}y + \dfrac{1}{2}x{y^2} - {y^4} + 1 + {x^3} + \dfrac{1}{2}{x^2}y - x{y^2} + {y^4} + 2\\
= 2{x^3} - \dfrac{3}{2}{x^2}y - \dfrac{1}{2}x{y^2} + 3\\
\to Bậc:3\\
c)E + F = 5xy - \dfrac{2}{3}{x^2}y + xy{z^2} - 1 + 2{x^2}y - xy{z^2} - \dfrac{2}{5}xy + x + \dfrac{1}{2}\\
= \dfrac{{23}}{5}xy + \dfrac{4}{3}{x^2}y - \dfrac{1}{2}\\
\to Bậc:3\\
E - F = 5xy - \dfrac{2}{3}{x^2}y + xy{z^2} - 1 - 2{x^2}y + xy{z^2} + \dfrac{2}{5}xy - x - \dfrac{1}{2}\\
= 2xy{z^2} + \dfrac{{27}}{5}xy - \dfrac{8}{3}{x^2}y - \dfrac{3}{2}\\
\to Bậc:4
\end{array}\)
