Câu hỏi của Đang Thuy Duyen

Question

Rut gon bieu thuc

A=(x+1)$\left(x^2+1\right)\left(x^4+1\right)...\left(x^{256}+1\right)+1$

Nguyễn Việt Lâm · Accepted Answer

$A-1=\left(x+1\right)\left(x^2+1\right)...\left(x^{256}+1\right)$

$\Rightarrow\left(A-1\right)\left(x-1\right)=\left(x-1\right)\left(x+1\right)\left(x^2+1\right)...\left(x^{256}+1\right)$

$\Rightarrow\left(A-1\right)\left(x-1\right)=\left(x^2-1\right)\left(x^2+1\right)...\left(x^{256}+1\right)$

$\Rightarrow\left(A-1\right)\left(x-1\right)=\left(x^4-1\right)\left(x^4+1\right)...\left(x^{256}+1\right)$

$\Rightarrow\left(A-1\right)\left(x-1\right)=\left(x^{256}-1\right)\left(x^{256}+1\right)=x^{512}-1$

$\Rightarrow A-1=\dfrac{x^{512}-1}{x-1}$

$\Rightarrow A=\dfrac{x^{512}-1}{x-1}+1=\dfrac{x^{512}+x-2}{x-1}$Câu trả lời: $A-1=\left(x+1\right)\left(x^2+1\right)...\left(x^{256}+1\right)$

$\Rightarrow\left(A-1\right)\left(x-1\right)=\left(x-1\right)\left(x+1\right)\left(x^2+1\right)...\left(x^{256}+1\right)$

$\Rightarrow\left(A-1\right)\left(x-1\right)=\left(x^2-1\right)\left(x^2+1\right)...\left(x^{256}+1\right)$

$\Rightarrow\left(A-1\right)\left(x-1\right)=\left(x^4-1\right)\left(x^4+1\right)...\left(x^{256}+1\right)$

$\Rightarrow\left(A-1\right)\left(x-1\right)=\left(x^{256}-1\right)\left(x^{256}+1\right)=x^{512}-1$

$\Rightarrow A-1=\dfrac{x^{512}-1}{x-1}$

$\Rightarrow A=\dfrac{x^{512}-1}{x-1}+1=\dfrac{x^{512}+x-2}{x-1}$

Violympic toán 8