簡約 (d^0+e^0)(d^0+e^0)

解

簡約

(d^{0} + e^{0}) (d^{0} + e^{0})

4

解答ステップ

(d^{0} + e^{0}) (d^{0} + e^{0})

指数の規則を適用する:

a^{b} \cdot a^{c} = a^{b + c}

(d^{0} + e^{0}) (d^{0} + e^{0}) = (d^{0} + e^{0})^{1 + 1}

= (d^{0} + e^{0})^{1 + 1}

数を足す:

1 + 1 = 2

= (d^{0} + e^{0})^{2}

規則を適用

a^{0} = 1, a \neq = 0

d^{0} = 1, e^{0} = 1

= (1 + 1)^{2}

数を足す:

1 + 1 = 2

= 2^{2}

2^{2} = 4

= 4