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nguyễn  xuân ly
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Nguyễn Lê Phước Thịnh
11 tháng 7 2021 lúc 14:30

a) Ta có: \(A=\dfrac{16^8-1}{\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)

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

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

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

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

\(=\dfrac{2^{32}-1}{2^{32}-1}=1\)

b) Ta có: \(B=\dfrac{\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{9^{16}-1}\)

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

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

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

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

nguyen hoang
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Trần Nguyễn Phương Thủy
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BLACK WHITE
4 tháng 11 2020 lúc 21:32

goi y nha A=1/2.(3^2-1)(3^2+1)....(3^32+1)

Khách vãng lai đã xóa
Đào Thị Khánh Vinh
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Thắng Nguyễn
7 tháng 7 2017 lúc 22:20

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

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

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

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

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

\(2A=3^{32}-1\Rightarrow A=\frac{3^{32}-1}{2}\)

Lục Tương
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Linh Kẹo
7 tháng 8 2016 lúc 18:18

mk chịu

Hoài Nam
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Nguyễn Lê Phước Thịnh
29 tháng 10 2021 lúc 20:51

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

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

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

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

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

....
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Yeutoanhoc
24 tháng 6 2021 lúc 10:15

`c)root{3}{4}.root{3}{1-sqrt3}.root{6}{(sqrt3+1)^2}`

`=root{3}{4(1-sqrt3)}.root{3}{1+sqrt3}`

`=root{3}{4(1-sqrt3)(1+sqrt3)}`

`=root{3}{4(1-3)}=-2`

`d)2/(root{3}{3}-1)-4/(root{9}-root{3}{3}+1)`

`=(2(root{3}{9}+root{3}{3}+1))/(3-1)-(4(root{3}{3}+1))/(3+1)`

`=root{3}{9}+root{3}{3}+1-root{3}{3}-1`

`=root{3}{9}`

Yeutoanhoc
24 tháng 6 2021 lúc 10:04

`a)root{3}{8sqrt5-16}.root{3}{8sqrt5+16}`

`=root{3}{(8sqrt5-16)(8sqrt5+16)}`

`=root{3}{320-256}`

`=root{3}{64}=4`

`b)root{3}{7-5sqrt2}-root{6}{8}`

`=root{3}{1-3.sqrt{2}+3.2.1-2sqrt2}-root{6}{(2)^3}`

`=root{3}{(1-sqrt2)^3}-sqrt2`

`=1-sqrt2-sqrt2=1-2sqrt2`

 

Khánh Bảo Trương
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Nguyễn Lê Phước Thịnh
29 tháng 7 2021 lúc 20:12

Ta có: \(\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

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

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

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

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

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

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

Khánh Nam.....!  ( IDΣΛ...
29 tháng 7 2021 lúc 20:14

Rút gọn: (3 + 1)(32 + 1)(34 + 1)(38 + 1)(316 + 1)(332 + 1)
A=(3 + 1)(32 + 1)(34 + 1)(38 + 1)(316 + 1)(332 + 1)

A=2(3 + 1)(32 + 1)(34 + 1)(38 + 1)(316 + 1)(332 + 1)

A=(3-1)(3 + 1)(32 + 1)(34 + 1)(38 + 1)(316 + 1)(332 + 1)
A=(32-1)(32 + 1)(34 + 1)(38 + 1)(316 + 1)(332 + 1)
A=(34-1)(34 + 1)(38 + 1)(316 + 1)(332 + 1)
A=(38-1)(38 + 1)(316 + 1)(332 + 1)
A=(316-1)(316 + 1)(332 + 1)
A=(332 - 1)(332 + 1)
A=364-1
=>A=(364-1) /2

Sakura Kinomoto
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