Câu 1:

\(A=\frac{2^5.7+2^5}{2^5.3-2^5}\)= \(\frac{2^5.8}{2^5.2}\)= 4

Vậy A = 4

Câu 2:

\(B=2^3.5^3-3.\left\{400-\left[673-2^3.\left(7^8:7^6+7^0\right)\right]\right\}\)

\(B=8.125-3.\left\{400-\left[673-8.\left(7^2+1\right)\right]\right\}\)

\(B=1000-3.\left\{400-\left[673-8.\left(49+1\right)\right]\right\}\)

\(B=1000-3.\left\{400-\left[673-8.50\right]\right\}\)

\(B=1000-3.\left\{400-\left[673-400\right]\right\}\)

\(B=1000-3.\left\{400-273\right\}\)

\(B=1000-3.127\)

\(B=1000-381\)

\(B=619\)

Vậy B = 619

1. x = 9/4

2. x = 7/3

3. x = 3/2 

Chúc bạn học tốt!!! ^^

sao đề kì quá sao mik ko tính ra hay là mik tính sai nhỉ! mik tính được ra là 77 nhưng thuer lại thì thấy kì 

\(x=\frac{11}{7}\)

=2011 do ban

cach giai la 

(1+2-3-4)+(5+6-7-8)+...............+(2005+2006-2007-2008)+2009+2010

=-4+-4+-4+4019

=(-4)x502+4019

=2008+4019

=2011

1, Không ai rảnh để làm cái này ra cho bạn. Mình nghĩ kết quả là 0.

2, Mình không có nick facebook.

Chúc bạn học tốt^^

ờ thế à pn có rảnh ko z

\(2^3\cdot5^3-3\left\{400-\left[673-2^3(7^8:7^6+7^0)\right]\right\}\)

\(=8\cdot125-3\left\{400-\left[673-8(7^{8-6}+7^0)\right]\right\}\)

\(=8\cdot125-3\left\{400-\left[673-8(7^2+7^0)\right]\right\}\)

\(=8\cdot125-3\left\{400-\left[673-8(49+1)\right]\right\}\)

\(=8\cdot125-3\left\{400-\left[673-8\cdot50\right]\right\}\)

\(=8\cdot125-3\left\{400-\left[673-400\right]\right\}\)

\(=8\cdot125-3\left\{400-273\right\}\)

\(=8\cdot125-3\cdot127\)

\(=619\)

\(2^3.5^3-3\left\{400-\left[673-2^3\left(7^8:7^6+7^0\right)\right]\right\}\)

\(=8.125-3\left\{400-\left[673-8\left(49+1\right)\right]\right\}\)

\(=1000-3\left\{400-\left[673-8.50\right]\right\}\)

\(=1000-3\left\{400-\left[673-400\right]\right\}\)

\(=1000-3\left\{400-273\right\}\)

\(=1000-3.\left(127\right)\)

\(=1000-381\)

\(=619\)

\(2^3.5^3-3\left\{400-\left[673-2^3\left(7^8:7^6+7^0\right)\right]\right\}\)

\(=8.125-\left\{400-\left[673-8\left(7^{8-6}+7^0\right)\right]\right\}\)

ta có 

\(S_2=\left(1-3\right)+\left(5-7\right)+..+\left(1997-1999\right)+2001\)

ha y \(S_2=-2-2-2..+2001=-2.500+2001=1001\)

\(S_3=\left(1-2-3+4\right)+\left(5-6-7+8\right)+..+\left(1997-1998-1999+2002\right)\)

hay \(S_3=0+0+..+0=0\)

\(S_2=\left(1-3\right)+\left(5-7\right)+...+\left(1997-1999\right)+2001\)

\(=\left(-2\right)+\left(-2\right)+....+\left(-2\right)+2001=\left(-2\right).500+2001=-1000+2001=1001\)

\(S_3=\left(0+1-2-3\right)+\left(4+5-6-7\right)+...+\left(1996+1997-1998-1999\right)+2000\)

\(=-4+\left(-4\right)+...+\left(-4\right)+2000=\left(-4\right).500+2000=0\)

2³ . 2⁵ - 3 { 400 - [ 673 - 2³ ( 7⁸ : 7⁶ +7⁰)]}

= 2⁸  -  3 { 400 - [ 673 - 8 ( 7² +1 )]}

= 256 - 3 { 400 - [ 673 - 8 . 50 ]}

= 256 - 3 { 400 - [ 673 - 400 ]}

= 256 - 3 { 400 - 273 }

= 256 - 3 . 127

= 256 - 381

= - 125

bằng 619 mới đúng nhé

C=1+2+3-4+5+6+7-8+...+1999-2000+2001

C=1+2+(-1)+5+6+(-1)+...+1997+1998+(-1)+2001

C=1+1+5+5+...+1997+1997+2001

C=2001+(1.2+5.2+...+1997.2)

C=2001+2(1+5+...+1997)

Tong 1+5+...+1997={(1997+1)[(1997-1):4+1]}:2=499500

C=2001+2.499500

C=2001+999000

C=1001001

he he