A=(3+1)(3^2+1)(3^4+1)(3^8+1) (không nhân ra nhé ) 03/12/2021 Bởi Eva A=(3+1)(3^2+1)(3^4+1)(3^8+1) (không nhân ra nhé )
Đáp án: $A = \dfrac{3^{16}-1}{2}$ Giải thích các bước giải: $\begin{array}{l}A = (3+1)(3^2 +1)(3^4+1)(3^8+1)\\ \to 2A = 2(3+1)(3^2 +1)(3^4+1)(3^8+1)\\ \to 2A = (3-1)(3+1)(3^2 +1)(3^4+1)(3^8+1)\\ \to 2A = (3^2-1)(3^2 +1)(3^4+1)(3^8+1)\\ \to 2A = (3^4-1)(3^4+1)(3^8+1)\\ \to 2A = (3^8-1)(3^8+1)\\ \to 2A = 3^{16}-1\\ \to A = \dfrac{3^{16}-1}{2} \end{array}$ Bình luận
`A=(3+1)(3^2+1)(3^4+1)(3^8+1)` `⇒2A=2(3+1)(3^2+1)(3^4+1)(3^8+1)` `⇒2A=(3-1)(3+1)(3^2+1)(3^4+1)(3^8+1)` `⇒2A=(3^2-1)(3^2+1)(3^4+1)(3^8+1)` `⇒2A=(3^4-1)(3^4+1)(3^8+1)` `⇒2A=(3^8-1)(3^8+1)` `⇒2A=3^(16)-1` `⇒A=(3^(16)-1)/2` Bình luận
Đáp án:
$A = \dfrac{3^{16}-1}{2}$
Giải thích các bước giải:
$\begin{array}{l}A = (3+1)(3^2 +1)(3^4+1)(3^8+1)\\ \to 2A = 2(3+1)(3^2 +1)(3^4+1)(3^8+1)\\ \to 2A = (3-1)(3+1)(3^2 +1)(3^4+1)(3^8+1)\\ \to 2A = (3^2-1)(3^2 +1)(3^4+1)(3^8+1)\\ \to 2A = (3^4-1)(3^4+1)(3^8+1)\\ \to 2A = (3^8-1)(3^8+1)\\ \to 2A = 3^{16}-1\\ \to A = \dfrac{3^{16}-1}{2} \end{array}$
`A=(3+1)(3^2+1)(3^4+1)(3^8+1)`
`⇒2A=2(3+1)(3^2+1)(3^4+1)(3^8+1)`
`⇒2A=(3-1)(3+1)(3^2+1)(3^4+1)(3^8+1)`
`⇒2A=(3^2-1)(3^2+1)(3^4+1)(3^8+1)`
`⇒2A=(3^4-1)(3^4+1)(3^8+1)`
`⇒2A=(3^8-1)(3^8+1)`
`⇒2A=3^(16)-1`
`⇒A=(3^(16)-1)/2`