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受欢迎的代数 >

化简 (4.3x^{105})left(2,1x^{103})left(6x^{10}-4)

解答

化简

(4.3 x^{105}) l e f t (2, 1 x^{103}) l e f t (6 x^{10} - 4)

解答

4.3 e^{2} l^{2} x^{105} f^{2} t^{2} (6 x^{10} - 4) (2 \cdot 1, x^{103})

求解步骤

(4.3 x^{105}) l e f t (2, 1 \cdot x^{103}) l e f t (6 x^{10} - 4)

使用指数法则:

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

ll = l^{1 + 1}

= 4.3 x^{105} e f t (2, 1 \cdot x^{103}) l^{1 + 1} e f t (6 x^{10} - 4)

数字相加:

1 + 1 = 2

= 4.3 x^{105} e f t (2, 1 \cdot x^{103}) l^{2} e f t (6 x^{10} - 4)

使用指数法则:

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

ee = e^{1 + 1}

= 4.3 x^{105} f t (2, 1 \cdot x^{103}) l^{2} e^{1 + 1} f t (6 x^{10} - 4)

数字相加:

1 + 1 = 2

= 4.3 x^{105} f t (2, 1 \cdot x^{103}) l^{2} e^{2} f t (6 x^{10} - 4)

使用指数法则:

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

ff = f^{1 + 1}

= 4.3 x^{105} t (2, 1 \cdot x^{103}) l^{2} e^{2} f^{1 + 1} t (6 x^{10} - 4)

数字相加:

1 + 1 = 2

= 4.3 x^{105} t (2, 1 \cdot x^{103}) l^{2} e^{2} f^{2} t (6 x^{10} - 4)

使用指数法则:

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

tt = t^{1 + 1}

= 4.3 x^{105} (2, 1 \cdot x^{103}) l^{2} e^{2} f^{2} t^{1 + 1} (6 x^{10} - 4)

数字相加:

1 + 1 = 2

= 4.3 x^{105} (2, 1 \cdot x^{103}) l^{2} e^{2} f^{2} t^{2} (6 x^{10} - 4)

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