展開する 4(sec^2(x^2+3x+2)(2x+3))

解

展開する

4 (sec^{2} (x^{2} + 3 x + 2) (2 x + 3))

8 x sec^{2} (x^{2} + 3 x + 2) + 12 sec^{2} (x^{2} + 3 x + 2)

解答ステップ

4 (sec^{2} (x^{2} + 3 x + 2) (2 x + 3))

括弧を削除する:

(a) = a

= 4 sec^{2} (x^{2} + 3 x + 2) (2 x + 3)

分配法則を適用する:

a (b + c) = ab + a c

a = 4 sec^{2} (x^{2} + 3 x + 2), b = 2 x, c = 3

= 4 sec^{2} (x^{2} + 3 x + 2) \cdot 2 x + 4 sec^{2} (x^{2} + 3 x + 2) \cdot 3

= 4 \cdot 2 x sec^{2} (x^{2} + 3 x + 2) + 4 \cdot 3 sec^{2} (x^{2} + 3 x + 2)

簡素化

4 \cdot 2 x sec^{2} (x^{2} + 3 x + 2) + 4 \cdot 3 sec^{2} (x^{2} + 3 x + 2) : 8 x sec^{2} (x^{2} + 3 x + 2) + 12 sec^{2} (x^{2} + 3 x + 2)

= 8 x sec^{2} (x^{2} + 3 x + 2) + 12 sec^{2} (x^{2} + 3 x + 2)