Đơn giản
\(C=\dfrac{2}{3}y^4-\dfrac{3}{4}y^2-\dfrac{9}{2}y-\dfrac{1}{3}y^3+y-\dfrac{2}{5}=-\dfrac{2}{5}-\dfrac{7}{2}y-\dfrac{3}{4}y^2-\dfrac{1}{3}y^3+\dfrac{2}{3}y^4\)\(D=\dfrac{1}{5}y^2-y-3y^4+2=2-y+\dfrac{1}{5}y^2-3y^4\)
Bây giờ tìm hiệu của C - D
Ta có hiệu của C - D bằng: ( xin lỗi vì không biết cách đặt tính )
\(C=-\dfrac{2}{5}-\dfrac{7}{2}y-\dfrac{3}{4}y^2-\dfrac{1}{3}y^3+\dfrac{2}{3}y^4-D=2-y+\dfrac{1}{5}y^2-3y^4\)
C - D = \(-\dfrac{12}{5}-\dfrac{5}{2}y-\dfrac{19}{20}y^2-\dfrac{1}{3}y^3+\dfrac{11}{3}y^4\)
\(C=\dfrac{2}{3}y^4-\dfrac{3}{4}\left(y+6\right)-\dfrac{1}{3}y^3+y-\dfrac{2}{5}\)
\(C=\dfrac{2}{3}y^4-\dfrac{3}{4}y-\dfrac{9}{2}-\dfrac{1}{3}y^3+y-\dfrac{2}{5}\)
\(C=\dfrac{3}{4}y^4-\dfrac{1}{3}y^3+\dfrac{1}{4}y-\dfrac{49}{10}\)
\(D=\dfrac{1}{5}y\left(y-5\right)-3y^4+2\)
\(D=\dfrac{1}{5}y^2-y-3y^4+2\)
\(D=-3y^4-\dfrac{1}{5}y^2-y+2\)
\(\Rightarrow C-D=\)\(\dfrac{3}{4}y^4-\dfrac{1}{3}y^3+\dfrac{1}{4}y-\dfrac{49}{10}-\)\(\left(-3y^4-\dfrac{1}{5}y^2-y+2\right)\)
\(=\dfrac{3}{4}y^4-\dfrac{1}{3}y^3+\dfrac{1}{4}y-\dfrac{49}{10}+3y^4+\dfrac{1}{5}y^2+y-2\)
\(=\dfrac{15}{4}y^4-\dfrac{1}{3}y^3+\dfrac{1}{5}y^2+\dfrac{5}{4}y-\dfrac{47}{10}\)
