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17 tháng 9 2018

A = 12 – 22 + 32 – 42 + … – 20042 + 20052

     A = 1 + (32 – 22) + (52 – 42)+ …+ ( 20052 – 20042)

     A = 1 + (3 + 2)(3 – 2) + (5 + 4 )(5 – 4) + … + (2005 + 2004)(2005 – 2004)

     A = 1 + 2 + 3 + 4 + 5 + … + 2004 + 2005

     A = ( 1 + 2002 ). 2005 : 2 = 2011015

b/  B = (2 + 1)(22 +1)(24 + 1)(28 + 1)(216 + 1)(232 + 1) – 264

     B = (22  - 1) (22 +1)(24 + 1)(28 + 1)(216 + 1)(232 + 1) – 264

     B = ( 24 – 1)(24 + 1)(28 + 1)(216 + 1)(232 + 1) – 264

     B = …

     B =(232 - 1)(232 + 1) – 264

     B = 264 – 1 – 264

     B = - 1

17 tháng 9 2018

xin lỗi nha chỗ câu a mình lộn

chỗ (1+2002)x2005:2=2011015 là sai nha 

       (1+2005)x2005:2= 2011015 là đúng nha 

b) Ta có: \(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)

\(=\left(2+1\right)\left(2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)

\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)

\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)

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

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

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

3 tháng 9 2019

b) \(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)

\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)

\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)

\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)

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

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

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

\(=-1\)

3 tháng 9 2019

\(\left(1^2-2^2\right)+\left(3^2-4^2\right)+....+\left(99^2-100^2\right)\) 

\(=\left(1-2\right)\left(2+1\right)+\left(3-4\right)\left(4+3\right)+....+\left(99-100\right)\left(100+99\right)\) 

\(=\left(-1\right)\left(1+2+3+....+100\right)=\frac{\left(-1\right)100.99}{2}=-4950\)

18 tháng 10 2015

Phân tích 3=4-1=\(2^2-1\)

10 tháng 8 2016

\(A=-1^2+2^2-3^2+4^2-...-99^2+100^2\)

\(=\left(2^2-1^2\right)+\left(4^2-3^2\right)+...+\left(100^2-99^2\right)\)

\(=\left(2+1\right)\left(2-1\right)+\left(4+3\right)\left(4-3\right)+...+\left(100+99\right)\left(100-99\right)\)

\(=1+2+3+4+...+99+100\)

\(=\frac{\left(1+100\right)\cdot100}{2}=5050\)

\(C=\left(2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)

\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)

\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)

\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)

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

\(=\left(2^{32}-1\right)\left(2^{32}+1\right)-2^{64}=\left(2^{64}-1\right)-2^{42}=-1\)

10 tháng 8 2016

Mk chỉ bt làm câu C thôi tại vì  mk chỉ học lớp 7leuleu

C=(2+1)(24+1)(28+1)(216+1)(232+1)-264

C=(24-1)(24+1)(28+1)(216+1)(232+1)-264

C=(28-1)(28+1)(216+1)(232+1)-264

C=(216-1)(216+1)(232+1)-264

C=(232-1)(232+1)-264

C=264-1-264

C=-1

17 tháng 8 2019

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

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

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

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

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

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

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

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

\(\Rightarrow B=2^{64}-1-2^{64}=-1\)

17 tháng 8 2019

Ta có : \(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)

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

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

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

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

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

Thay 264 - 1 vào B, ta được :

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

14 tháng 7 2015

Câu b đúng r mà trieu dang

13 tháng 7 2015

như thế này chứ:

A=1002-992+982-972+...+22-12

B=12-22+32-42+...-20082-20092

C=3.(22+1)(24+1)(28+1)(216+1)-232

26 tháng 10 2018

Bài 1:

a) \(100^2-99^2+...+2^2-1^2\)

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

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

=> tự làm tiếp :))

b) tương tự

Bài 2 :

a) \(A=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)

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

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

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

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

\(A=2^{16}-1< 2^6=B\)

b) Phân tích \(2004\cdot2006=\left(2005-1\right)\left(2005+1\right)=\left(2005^2-1\right)\)rồi áp dụng hđt thứ 3 tự làm tiếp như câu a)

Bài 3:

a) Cứ khai triển hết ra 

b) \(a^2+b^2+c^2=ab+bc+ac\)

\(a^2+b^2+c^2-ab-bc-ac=0\)

Nhân 2 vào cả 2 vế được :

\(2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)

\(\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ac+c^2\right)=0\)

\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

mà mũ 2 luôn lớn hơn hoặc bằng 0

\(\Rightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}\Rightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}\Rightarrow}a=b=c\left(đpcm\right)}\)

P.s: toàn bài nâng cao làm hơi ẩu tí ^^