Bài 3: Những hằng đẳng thức đáng nhớ

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ĐỖ THỊ THANH HẬU

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

Giúp tớ tí nhenlolang

Hà Linh
13 tháng 7 2017 lúc 8:05

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

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

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

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

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

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

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

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

2A = \(3^{64}-1\)

A = \(\dfrac{3^{64}-1}{2}\)

Nguyễn Thị Huyền Trang
13 tháng 7 2017 lúc 8:07

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

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

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

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

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

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

qwerty
13 tháng 7 2017 lúc 8:08

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

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

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

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

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

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

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

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


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