khong biet

\(\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

\(=\frac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

\(=\frac{1}{2}\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

\(=\frac{1}{2}\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

\(=\frac{1}{2}\left(3^{16}-1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

\(=\frac{1}{2}\left(3^{32}-1\right)\left(3^{32}+1\right)\)

\(=\frac{1}{2}\left(3^{64}-1\right)\)

Ai ngu mới trả lời câu này

dung vay có ngu mới tra lơi

a) \(x^8+x+1\)

\(=x^8+x^7+x^6-x^7-x^6-x^5+x^5+x^4+x^3-x^4-x^3-x^2+x^2+x+1\)

\(=x^6\left(x^2+x+1\right)-x^5\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)-x^2\left(x^2+x+1\right)+\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^6-x^5+x^3-x^2+1\right)\)

b) \(x^4+2008x^2+2007x+2008\)

\(=x^4+x^3+x^2-x^3-x^2-x+2008x^2+2008x+2008\)

\(=x^2\left(x^2+x+1\right)-x\left(x^2+x+1\right)+2008\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+2008\right)\)

\(x^8+x+1\)

\(=x^8+x^7+x^6-x^7-x^6-x^5+x^5+x^4+x^3-x^4-x^3-x^2+x^2+x+1\)

\(=x^6\left(x^2+x+1\right)-x^5\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)-x^2\left(x^2+x+1\right)+x^2+x+1\)

\(=\left(x^2+x+1\right)\left(x^6-x^5+x^3-x^2+1\right)\)

 Tao ko thể ngờ là mày ngu đến vậy Hoàng Phúc Tao là Nguyễn Việt Hà