簡約 (3(\sqrt[3]{x+1})+4)^2

解

簡約

(3 (3 x + 1) + 4)^{2}

9 (3 x + 1)^{2} + 24 3 x + 1 + 16

解答ステップ

(3 (3 x + 1) + 4)^{2}

規則を適用する:

(a) = a

(3 x + 1) = 3 x + 1

= (3 3 x + 1 + 4)^{2}

完全平方式を適用する:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

(3 3 x + 1 + 4)^{2} = (3 3 x + 1)^{2} + 2 \cdot 3 3 x + 1 \cdot 4 + 4^{2}

= (3 3 x + 1)^{2} + 2 \cdot 3 3 x + 1 \cdot 4 + 4^{2}

2 \cdot 3 3 x + 1 \cdot 4 = 24 3 x + 1

4^{2} = 16

= (3 3 x + 1)^{2} + 24 3 x + 1 + 16

指数の規則を適用する:

(a \cdot b)^{n} = a^{n} b^{n}

= 3^{2} (3 x + 1)^{2} + 24 3 x + 1 + 16

3^{2} = 9

= 9 (3 x + 1)^{2} + 24 3 x + 1 + 16