展開する (2+x)[\H(x-2)-H[x-3]]yy_{2}

解

展開する

(2 + x) [H (x - 2) - H [x - 3]] y y_{2}

- x^{2} Hy y_{2} + x Hy y_{2} + x y y_{2} H (x - 2) + 2 y y_{2} H (x - 2) + 6 Hy y_{2}

解答ステップ

(2 + x) [H (x - 2) - H [x - 3]] y y_{2}

= y y_{2} (2 + x) [H (x - 2) - H [x - 3]]

分配法則を適用する:

a (b + c) = ab + a c

a = [H (x - 2) - H [x - 3]] y y_{2}, b = 2, c = x

= [H (x - 2) - H [x - 3]] y y_{2} \cdot 2 + [H (x - 2) - H [x - 3]] y y_{2} x

= 2 y y_{2} [H (x - 2) - H [x - 3]] + x y y_{2} [H (x - 2) - H [x - 3]]

簡素化

2 y y_{2} [H (x - 2) - H [x - 3]] + x y y_{2} [H (x - 2) - H [x - 3]] : - x^{2} Hy y_{2} + x Hy y_{2} + x y y_{2} H (x - 2) + 2 y y_{2} H (x - 2) + 6 Hy y_{2}

= - x^{2} Hy y_{2} + x Hy y_{2} + x y y_{2} H (x - 2) + 2 y y_{2} H (x - 2) + 6 Hy y_{2}