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22 tháng 10 2020

Đặt A = ( 2 + 1 )( 22 + 1 )( 24 + 1 )...( 2256 + 1 )

=> ( 2 - 1 )A = ( 2 - 1 )( 2 + 1 )( 22 + 1 )( 24 + 1 )...( 2256 + 1 )

=> A = ( 22 - 1 )( 22 + 1 )( 24 + 1 )...( 2256 + 1 )

         = ( 24 - 1 )( 24 + 1 )...( 2256 + 1 )

         = ( 2256 - 1 )( 2256 + 1 )

         = 2512 - 1

20 tháng 9 2021

\(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{10}+1\right)+1\)

\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)...\left(2^{20}+1\right)+1\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{20}+1\right)+1\)

\(=2^{40}-1+1=2^{40}\)

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

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

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

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

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

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

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

AH
Akai Haruma
Giáo viên
23 tháng 6 2023

1. 

$=153^2+2.47.153+47^2=(153+47)^2=200^2=40000$

2.

$=1,24^2-2.1,24.0,24+0,24^2=(1,24-0,24)^2=1^2=1$

3. Không phù hợp để tính nhanh 

4. 

$=15^8-(15^8-1)=1$

5.

$=(1^2-2^2)+(3^2-4^2)+(5^2-6^2)+...+(2019^2-2020^2)$

$=(1-2)(1+2)+(3-4)(3+4)+(5-6)(5+6)+...+(2019-2020)(2019+2020)$

$=(-1)(1+2)+(-1)(3+4)+(-1)(5+6)+....+(-1)(2019+2020)$

$=(-1)(1+2+3+4+....+2019+2020)=(-1).2020(2020+1):2=-2041210$

DT
23 tháng 6 2023

6:

\(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)....\left(2^{2020}+1\right)+1\\ =1.\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)....\left(2^{2020}+1\right)+1\\ =\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)....\left(2^{2020}+1\right)+1\\ =\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)....\left(2^{2020}+1\right)+1\\ =\left(2^4-1\right)\left(2^4+1\right)....\left(2^{2020}+1\right)+1\\ =\left(2^8-1\right)....\left(2^{2020}+1\right)+1\\ =\left(2^{2020}-1\right)\left(2^{2020}+1\right)+1\\ =2^{4040}-1+1=2^{4040}\)

29 tháng 3 2019

Ta có

N   =   ( 2   +   1 ) ( 2 2   +   1 ) ( 2 4   +   1 ) ( 2 8   +   1 ) ( 2 16   +   1 )     ( 2 16   +   1 )   =   3 ( 2 2   +   1 ) ( 2 4   +   1 ) ( 2 8   +   1 )     ( 2 16   +   1 )   =   [ ( 2 2   –   1 ) ( 2 2   +   1 ) ] ( 2 4   +   1 ) ( 2 8   +   1 ) ( 2 16   +   1 )     =   ( 2 4   –   1 ) ( 2 4   +   1 ) ( 2 8   +   1 ) ( 2 16   +   1 )     =   ( 2 8   –   1 ) ( 2 8   +   1 ) ( 2 16   +   1 )     =   ( 2 16   -   1 ) ( 2 16   +   1 )   = 2 16 2 − 1 = 2 32 − 1 M à   2 32 − 1 > 2 32 ⇒   N < M

Đáp án cần chọn là: A

10 tháng 12 2023

1,

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

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

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

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

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

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

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

Vậy \(A=2^{32}-1\)

2, \(x^2-6x=-9\)

\(x^2-6x+9=0\)

\(\left(x-3\right)^2=0\)

\(x-3=0\)

\(x=3\)

Vậy \(x=3\)

a: \(A=\left(100-99\right)\left(100+99\right)+\left(98+97\right)\left(98-97\right)+....+\left(2+1\right)\left(2-1\right)\)

\(=100+99+98+97+...+2+1\)

=5050

b: \(B=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\cdot...\cdot\left(2^{64}+1\right)+1\)

\(=\left(2^4-1\right)\left(2^4+1\right)\cdot...\cdot\left(2^{64}+1\right)+1\)

\(=\left(2^8-1\right)\left(2^8+1\right)\cdot...\cdot\left(2^{64}+1\right)+1\)

\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)+1\)

\(=\left(2^{32}-1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)+1\)

\(=\left(2^{64}-1\right)\cdot\left(2^{64}+1\right)+1\)

\(=2^{128}-1+1=2^{128}\)

20 tháng 2 2022

a. \(A=100^2-99^2+98^2-97^2+...+2^2-1^2\)

\(=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+...+\left(2-1\right)\left(2+1\right)\)

\(=199+195+...+3\)

\(=\dfrac{\left(199+3\right)\left(\dfrac{199-3}{4}+1\right)}{2}=5050\)

b. \(B=3\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1^2\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1^2\)

\(=\left(2^4-1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1^2\)

\(=2^{128}-1+1=2^{128}\)

c) \(C=\left(a+b+c\right)^2+\left(a+b-c\right)^2-2\left(a+b\right)^2\)

\(=a^2+b^2+c^2+2ab+2ac+2bc+a^2+b^2+c^2+2ab-2ac-2bc-2a^2-2b^2-4ab\)

\(=2c^2\)

20 tháng 10 2021

các bn giúp mình nhé

20 tháng 10 2021

chụp khó nhìn quá bn ơi

a: A=(100-99)(100+99)+(98-97)(98+97)+...+(2-1)(2+1)

=100+99+98+...+2+1

=5050

b: \(B=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\cdot...\cdot\left(2^{64}+1\right)+1\)

\(=\left(2^4-1\right)\left(2^4+1\right)\cdot...\cdot\left(2^{64}+1\right)+1\)

\(=\left(2^8-1\right)\left(2^8+1\right)\cdot...\cdot\left(2^{64}+1\right)\)+1

\(=2^{64}-1+1=2^{64}\)

19 tháng 6 2021

`A=(2-1)(2+1)(2^2+1)...(2^16+1)`

`=(2^2-1)(2^2+1)....(2^16+1)`

`=(2^4-1)....(2^16+1)`

`=2^32-1<2^32`

`=>A<B`