1/A=2004.2006 B=$2005^2$2/A=$\left(3 1\right)\left(3^2 1\right)\left(3^4 1\right)\left(3^8 1\right)\left(3^{16} 1\right)$va B=$3^{22}-1$3/ A=2004.2006.$2008^2$vaa...

0

NN

Nguyễn Ngọc k10

7 tháng 7 2023

BT7: Tính

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

$2,B=\left(1-3\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{16}+1\right)$

2

NL

Nguyễn Lê Phước Thịnh

7 tháng 7 2023

1: A=(3^2-1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)

=(3^4-1)(3^4+1)(3^8+1)(3^16+1)

=(3^8-1)(3^8+1)(3^16+1)

=(3^16-1)(3^16+1)

=3^32-1

2: B=(1-3^2)(1+3^2)*...*(1+3^16)

=(1-3^4)(1+3^4)(1+3^8)(1+3^16)

=1-3^32

Đúng(1)

GH

Gia Huy

7 tháng 7 2023

1

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

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

Đúng(2)

LT

LÊ TRẦN BÁCH

VIP

17 tháng 11 2023

...

#Toán lớp 6

2

NT

Nguyễn Thị Thương Hoài

Giáo viên VIP

18 tháng 11 2023

A = - 5²² - { - 222 - [ - 122 - (100 - 5²²) + 2022] }

A = - 5²² - { -222 - [- 122 - 100 + 5²² ] + 2022}

A = - 5²² - { -222 - { - 222 + 5²² } + 2022}

A = - 5²² - {- 222 + 222 - 5²² + 2022}

A = -5²² + 5²² - 2022

A = - 2022

Đúng(1)

NT

Nguyễn Thị Thương Hoài

Giáo viên VIP

18 tháng 11 2023

B = 1 + $\dfrac{1}{2}$(1 + 2) + $\dfrac{1}{3}$.(1 + 2 + 3) + ... + $\dfrac{1}{20}$.(1 + 2+ 3 + ... + 20)

B = 1+$\dfrac{1}{2}$$\times$(1+2)$\times$[(2-1):1+1]:2+ ... + $\dfrac{1}{20}$$\times$ (20 + 1)$\times$[(20-1):1+1]:2

B = 1 + $\dfrac{1}{2}$ $\times$ 3 $\times$ 2:2 + $\dfrac{1}{3}$ $\times$4 $\times$ 3 : 2+....+ $\dfrac{1}{20}$ $\times$21 $\times$ 20 : 2

B = 1 + $\dfrac{3}{2}$ + $\dfrac{4}{2}$ + ....+ $\dfrac{21}{2}$

B = $\dfrac{2+3+4+...+21}{2}$

B = $\dfrac{\left(21+2\right)\left[\left(21-2\right):1+1\right]:2}{2}$

B = $\dfrac{23\times20:2}{2}$

B = $\dfrac{23\times10}{2}$

B = 23

HD

Hòa Đình

18 tháng 6 2018

rút gọn biểu thức

a, $\left(a-b+c\right)^2-\left(b-c\right)^2+2ab-2ac$

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

1

TV

Trần Việt Linh

18 tháng 6 2018

a) $\left(a-b+c\right)^2-\left(b-c\right)^2+2ab-2ac$

$=\left(a^2+\left(-b\right)^2+c^2-2ab+2ac-2bc\right)-\left(b^2-2bc+c^2\right)+2ab-2ac$

$=a^2+b^2+c^2-2ab+2ac-2bc-b^2+2bc-c^2+2ab-2ac$

$=a^2+b^2-b^2+c^2-c^2-2ab+2ab+2ac-2ac-2bc+2bc$

$=a^2$

TY

Thiên Yết

21 tháng 5 2017

Tính nhanh:
a) A=$1^2-2^2+3^2-4^2+...-2008^2+2009^2$

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

1

KB

Kẻ Bí Mật

21 tháng 5 2017

b, B=(2+1)(2²+1)(2⁴+1)(2⁸+1)(2¹⁶+1)-2³²

=(2⁴-1)(2⁴+1)(2⁸+1)(2¹⁶+1)-2³²

=(2⁸-1)(2⁸+1)(2¹⁶+1)-2³²

=(2^16-1)(2¹⁶+1)-2³²

=2³²-1-2³²

=-1

VD

Vô Danh

24 tháng 8 2017

Gửi các bạn bài toán tính nhanh nè

a, A=$\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{6}\right).\left(1+\dfrac{1}{10}\right).\left(1+\dfrac{1}{15}\right).\left(1-\dfrac{1}{21}\right).\left(1-\dfrac{1}{28}\right)$

d, D=$\dfrac{2006.2005-1}{2004.2006+2005}$

XEM BẠN NÀO GIẢI DC TRƯỚC NÈ

#Toán lớp 7

1

MP

Mysterious Person

24 tháng 8 2017

mk chỉ giải đc câu d thôi nha ; bn thông cảm

d) $D=\dfrac{2006.2005-1}{2004.2006+2005}=\dfrac{2006.2005-1}{2004.2006+2006-1}$

$D=\dfrac{2006.2005-1}{2006\left(2004+1\right)-1}=\dfrac{2006.2005-1}{2006.2005-1}=1$

rút gọn biểu thức a/ 2x$\left(2x-1\right)^2-3x\left(x+3\right)\left(x-3\right)-4x\left(x+1\right)^2$ b/ $\left(a-b+c\right)^2-\left(b-c\right)^2+2ab-2ac$ c/ $\left(3x+1\right)^2-2\left(3x+1\right)\left(3x+5\right)+\left(3x+5\right)^2$ d/ ( 3 + 1 )...

L

Linh

12 tháng 10 2017

1

TT

Trịnh Thùy DƯơng

25 tháng 8 2018

rút gọn biểu thức

$a)$ 2x $(2x-1) 2 -3x(x+3)(x-3)-4x(x+1) 2$

=2x(4x²-4x+1)-3x.(x²-9)-4x(x²+2x+1)

=8x³-8x²+2x-3x³-27x-4x³-8x²-4x

=8x³-16x²-7x³-29x

Rút gọn : \(a,A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\\ b,B=-1^2+2^2-3^2+4^2-...-99^2+100^2\\ c,C=-1^2+2^2-3^2+4^2-...+\left(-1\right)^n\cdot...

IJ

I ♥ Jungkook

25 tháng 7 2018

5

AT

Aki Tsuki

24 tháng 8 2018

nhiều thế, đăng ít một thôi bạn

AT

Aki Tsuki

24 tháng 8 2018

a/ $A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)$

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

$2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)$

$2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)$

$2A=\left(3^4-1\right)\left(3^4+1\right)...\left(3^{64}+1\right)$

$\Rightarrow2A=3^{128}-1\Rightarrow A=\dfrac{3^{128}-1}{2}$

DT

Duong Thi Nhuong

18 tháng 4 2017

Tính rồi so A và B :

$A=\left(0,25\right)^{-1}.\left(1\dfrac{1}{4}\right)^2+25\left[\left(\dfrac{4}{3}\right)^{-2}:\left(1,25\right)^3\right]:\left(\dfrac{-2}{3}\right)^{-3}$

$B=\left(0,2\right)^{-3}.\left[\left(\dfrac{-1}{5}\right)^{-2}\right]^{-1}+\left[\left(\dfrac{1}{2}\right)^{-3}\right]^{-2}:\left(\dfrac{1}{8}\right)^{-1}-\left(2^{-3}\right)^{-2}:\dfrac{1}{2^6}$

1

NN

Ngáo Nu

19 tháng 4 2017

$A=4.\dfrac{25}{16}+25.\left[\dfrac{9}{16}:\dfrac{125}{64}\right]:\dfrac{-27}{8}$

$=\dfrac{25}{16}+25.\dfrac{36}{125}:\dfrac{-27}{8}=-\dfrac{137}{240}\left(1\right)$

$B=125.\left[\dfrac{1}{25}+\dfrac{1}{64}:8\right]-64.\dfrac{1}{64}$

$=125.\dfrac{89}{1600}:8-64.\dfrac{1}{64}=\dfrac{-67}{512}\left(2\right)$

Vì (2) > (1) => B > A

NK

Nguyễn Khánh Ly

17 tháng 3 2017

So sánh A và B biết

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

$B=3^{32}$

2

TT

Trần Thiên Kim

18 tháng 3 2017

Nếu đề thế này thì mình có thể làm được:

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

$\Rightarrow2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)$

$\Rightarrow2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)$

$\Rightarrow2A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)$

$\Rightarrow2A=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)$

$\Rightarrow2A=\left(3^{16}-1\right)\left(3^{16}+1\right)$

$\Rightarrow2A=3^{32}-1$

$\Rightarrow A=\dfrac{3^{32}-1}{2}$

=> B>A

DC

dam cong tian

17 tháng 3 2017

B>A