Tính
(72009+72010):72009
23.52-42.10+130.10000
290.150-53.2+5.11000
(22008+152007).(39+22007).(24-42)
Cho A = 1 + 2 + 2 2 + . . . + 2 2007 . Chứng minh: A = 2 2008 - 1
Cho A = 1 + 2 + 2 2 + . . . + 2 2007 . Chứng minh: A = 2 2008 - 1
A = 1 + 2 + 2 2 + . . . + 2 2007
2 A = 2 + 2 2 + . . . + 2 2007 + 2 2008
A = 2A - A = ( 2 + 2 2 + . . . + 2 2007 + 2 2008 ) - ( 1 + 2 + 2 2 + . . . + 2 2007 ) = 2 2008 - 1
Vậy A = 2 2008 - 1
42 . 2
24.2
42.10
tích mik mik tích lại
84
48
420
tích mik mik tích lại
mik ko nói xạo
Cho biểu thức:
A = 2009 . 2010 - 2 2008 + 2008 . 2010 ; B = - 2009 . 20102010 20092009 . 2010
Tính A+B
a) Tính M = 22010 - ( 22009 + 22008 + ..... + 21 + 20 )
b) So sánh: 2332 và 3223
Cho A = 1+21+22+233+...+22007
a)Tính 3A
b)Chứng minh : A = 22008--1
A \(=\)\(1+2^1+2^2+...+2^{2007}\)
⇒2 A \(=\)\(2+2^2+...+2^{2007}+2^{2008}\)
2A - A \(=\)( \(2+2^2+...+2^{2007}+2^{2008}\) ) - ( \(1+2^1+2^2+...+2^{2007}\) )
A\(=\)\(2^{2008}-1\)
\(3A=3\left(2^{2008}-1\right)\)
\(=3.2^{2008}-3\)
Tính các tổng sau
a, A = 1 + 5 3 + 5 5 + 5 7 + . . . + 5 99
b, A = 1 - 2 + 2 2 - . . . - 2 2007
c, A = 7 + 7 3 + 7 5 + 7 7 + . . . + 7 1999
a, A = 1 + 5 3 + 5 5 + 5 7 + . . . + 5 99
B = 5 4 + 5 6 + 5 8 + . . . + 5 100 = 5 . ( 5 3 + 5 5 + 5 7 + . . . + 5 99 ) = 5(A – 1)
A + B – 1 = 5 3 + 5 4 + . . . + 5 100
5(A + B – 1) = 5 4 + 5 5 + . . . + 5 100 + 5 101
4(A + B – 1) = 5(A + B – 1) – (A + B – 1) = 5 101 - 5 3
=> A + B – 1 = 5 101 - 5 3 4
=> A + 5(A – 1) –1 = 5 101 - 5 3 4 => 6A – 6 = 5 101 - 5 3 4
=> A – 1 = 5 101 - 5 3 24
=> A = 5 101 - 5 3 + 24 24
b, A = 1 - 2 + 2 2 - . . . - 2 2007
A = 1 + 2 2 + . . . + 2 2006 - 2 + 2 3 + . . . + 2 2007
A = ( 1 + 2 2 + . . . + 2 2006 ) - 2 . 1 + 2 2 + . . . + 2 2006
A = - 1 + 2 2 + . . . + 2 2006
Đặt B = - 2 + 2 3 + . . . + 2 2007 = - 2 . 1 + 2 2 + . . . + 2 2006 = 2A
A + B = - 1 + 2 + 2 2 + . . . + 2 2006 + 2 2007
2(A+B) = - 2 + 2 2 + . . . + 2 2006 + 2 2007 + 2 2008
A+B = 2(A+B)–(A+B) = - 2 2008 - 1
=> A+2A = - 2 2008 - 1
=> 3A = - 2 2008 - 1
=> A = - ( 2 2008 - 1 ) 3
c, A = 7 + 7 3 + 7 5 + 7 7 + . . . + 7 1999
Đặt B = 7 2 + 7 4 + 7 6 + . . . + 7 1999 + 7 2000 = 7 ( 7 + 7 3 + 7 5 + 7 7 + . . . + 7 1999 ) = 7A
A+B = 7 + 7 2 + 7 3 + . . . + 7 1999 + 7 2000
7(A+B) = 7 2 + 7 3 + . . . + 7 1999 + 7 2000 + 7 2001
7(A+B) – (A+B) = ( 7 2 + 7 3 + . . . + 7 1999 + 7 2000 + 7 2001 ) – ( 7 + 7 2 + 7 3 + . . . + 7 1999 + 7 2000 )
6(A+B) = 7 2001 - 7
A+B = 7 2001 - 7 6
=> A + 7A = 7 2001 - 7 6 => 8A = 7 2001 - 7 6 => A = 7 2001 - 7 48
Tính các tổng sau:
a) A = 1 + 5 3 + 5 5 + 5 7 + . . . + 5 99
b) A = 1 - 2 + 2 2 - . . . - 2 2007
c) A = 7 + 7 3 + 7 5 + 7 7 + . . . + 7 1999
8 :' x , 12 :' x , 24 :' x
b) 24 :' x , 30 :' x
c) 42 :' x , 58 :' x
giúp e
8 ⋮ \(x\); 12 ⋮ \(x\); 24 ⋮ \(x\) ⇒ 8; 12; 24 ⋮ \(x\) ⇒ \(x\) \(\in\) ƯC(8;12;24)
8 = 23; 12 = 22.3; 24 = 22.3
ƯCLN(8; 12; 24) = 22 = 4
\(x\) \(\in\) {-4; -2; -1; 1; 2; 4}
b, 24 ⋮ \(x\); 30 ⋮ \(x\) ⇒ \(x\) \(\in\) ƯC(24; 30)
24 = 23.3; 30 = 2.3.5 ⇒ ƯCLN(24; 30) = 2.3 = 6
\(x\) \(\in\) Ư(6) = {-6; -3; -2; -1; 1; 2; 3; 6}
tính 5,32*5959+0.532*40+53.2*203.3+532*20.04
=5.32x5959+0.532x10x4+5.32x10x203.3+5.32x100x20.04
=5.32x5959+5.32x4+5.32x2003+5.32x2004
=5.32x(5959+4+2003+2004)
=5.32x9970=53040.4