因式分解 36^{2018}+36^{2019}

解答

因式分解

3 6^{2018} + 3 6^{2019}

6^{4036} \cdot 37

求解步骤

3 6^{2018} + 3 6^{2019}

36 = 2^{2} \cdot 3^{2}

= (2^{2})^{2018} (3^{2})^{2018} + (2^{2} \cdot 3^{2})^{2019}

使用指数法则:

(ab)^{c} = a^{c} b^{c}

= (2^{2})^{2018} (3^{2})^{2018} + (2^{2})^{2019} (3^{2})^{2019}

使用指数法则:

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

2^{2019 \cdot 2} = 2^{2018 \cdot 2} \cdot 2^{2}, 3^{2019 \cdot 2} = 3^{2018 \cdot 2} \cdot 3^{2}

= 2^{2018 \cdot 2} \cdot 3^{2018 \cdot 2} + 2^{2018 \cdot 2} \cdot 2^{2} \cdot 3^{2018 \cdot 2} \cdot 3^{2}

因式分解出通项

3^{2018 \cdot 2} \cdot 2^{2018 \cdot 2}

= 3^{2018 \cdot 2} \cdot 2^{2018 \cdot 2} (1 + 2^{2} \cdot 3^{2})

整理后得

= 6^{4036} \cdot 37

f a c t or (- 67.4) + (+ 51.3 p)

f a c t or l (e^{4} t cos (t))

f a c t or e^{34704} + e^{3555} \cdot 7

f a c t or 9^{2} - 30 x + 25

f a c t or - 200