bài 6:tính nhanh
7)1\(^2\)-2\(^2\)+3\(^2\)-4\(^2\)+....-2004\(^2\)+2005\(^2\)
8) (2+1)(2\(^2\)+1)(2\(^4\)+1)(2\(^8\)+1)(2\(^{16}\)+1)(2\(^{32}\)+1)-2\(^{64}\)
tính : 1^2-2^2+3^2-4^2+...-2004^2+2005^2
b) (2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)-2^64
A = 12 – 22 + 32 – 42 + … – 20042 + 20052
b/ B = (2 + 1)(22 +1)(24 + 1)(28 + 1)(216 + 1)(232 + 1) – 264
giúp nhe ai nhanh mình tk
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
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
/ 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 + 2005 ). 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
đây là bài hoàn chình nè
1)_ Tìm phân số, biết rằng nếu lấy 3 / 2 trừ đi phân số đó rồi cộng với 5 / 7 thì được kết quả là 11 / 14
2)_ Tính hợp lý :
a. 17 / 9 + 19 / 13 + 14 / 6 + 7 / 13 + 10 / 6 + 1 / 9
b. 2005 x 2007 - 1 / 2004 + 2005 x 2006
3)_ Tính tổng :
A = 1 / 2 + 1 / 6 + 1 / 12 + . . . + 1 / 42
B = 2 / 1 x 2 + 2 / 2 x 3 + 2 / 3 x 4 + . . . + 2 / 8 x 9
C = 1 / 2 + 1 / 4 + 1 / 8 + 1 / 16 + 1 / 32 + 1 / 64
4)_ Tìm y :
a. y x 5 / 6 = 1 + 1 / 2
b. 15 / 28 : y = 1 / 7 x 4
c. 42 / 25 : y / 5 = 1 : 5 / 6
a/b nhân 4 cộng 1/6 = 17/6 số phải tìm là bao nhiêu
Mỗi bài mình làm một dạng thôi nhé!
1) \(\left(\frac{3}{2}-\frac{x}{y}\right)+\frac{5}{7}=\frac{11}{14}\)
\(\Rightarrow\frac{x}{y}=\frac{3}{2}-\left(\frac{11}{14}-\frac{5}{7}\right)=\frac{10}{7}\)
2) a)
\(\frac{17}{9}+\frac{19}{13}+\frac{14}{6}+\frac{7}{13}+\frac{10}{6}+\frac{1}{9}\)
\(=\left(\frac{17}{9}+\frac{1}{9}\right)+\left(\frac{19}{13}+\frac{7}{13}\right)+\left(\frac{14}{6}+\frac{10}{6}\right)\)
\(=2+2+4\)
\(=8\)
3) a)
\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{42}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{6.7}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{6}-\frac{1}{7}\)
\(A=\frac{1}{1}-\frac{1}{7}=\frac{6}{7}\)
4) a)
\(y.\frac{5}{6}=1+\frac{1}{2}\)
\(\Rightarrow y.\frac{5}{6}=\frac{3}{2}\)
\(\Rightarrow y=\frac{3}{2}.\frac{6}{5}=\frac{9}{5}\)
Tính nhanh
A. 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128
B. 2/1*3 + 2/3*5 + 2/5*7 + 2/7*9 + 2/9*11
Bài 1 Tính nhanh
a) -3/7 + 5/13 + 3/7
b) -5/21+-2/21+8/24
c) -5/11+(-6/11+2)
d) (-1/32+1/2)+15/32
e)5/17+ -6/13 + 3/4 + 7/-13+12/17
f) 7/23+-18/18+-4/9+16/23+-5/8
g)1/3+-3/4+3/5+-1/36+1/15+-2/9
h)-1/2+1/3+-1/4+-2/8+4/18+4/9
a)\(-\dfrac{3}{7}+\dfrac{5}{13}+\dfrac{3}{7}\)
=\(\left(-\dfrac{3}{7}+\dfrac{3}{7}\right)+\dfrac{5}{13}\)
=\(0+\dfrac{5}{13}\)
=\(\dfrac{5}{13}\)
BÀI TÍNH NHANH
A.2/3+2/6+2/12+2/24+2/48+2/96+2/192
B.1/2+1/4+1/8+1/16+1/32+1/64+1/128+1/256
C.1/3+1/9+1/27+1/81+1/243+1/729
D.3/2+3/8+3/32+3/128+3/512
E.3+3/5+3/25+3/125+3/625
a) = \(\frac{127}{96}\)
b) = \(\frac{255}{256}\)
c) Mik bỏ nha
d) = \(\frac{1023}{512}\)
e) = \(\frac{2343}{625}\)
Tính:
a) A= -1^2+2^2-3^2+4^2-...-99^2+100^2
b) B= 1-2^2+3^2-4^2+...-2004^2+2005^2
c) C= (2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)-2^64
THANKS
a,A=-(12-22+32-42+...+992-1002)
=-[(1-2)(1+2)+(3-4)(3+4)+...+(99-100)(99+100)]
=-[(-1).3+(-1).7+...+(-1).199]
=-[(-1).(3+7+...+199]
=\(\frac{\left(199+3\right).50}{2}=5050\)
b, tương tự a
c) C=1(2+1)(22+1)(24+1)(28+1)(216+1)(232+1)-264
=(2-1)(2+1)(22+1)(24+1)(28+1)(216+1)(232+1)-264
=(22-1)(22+1)(24+1)(28+1)(216+1)(232+1)-264
=(24-1)(24+1)(28+1)(216+1)(232+1)-264
=(28-1)(28+1)(216+1)(232+1)-264
=(216-1)(216+1)(232+1)-264
=(232-1)(232+1)-264
=264-1-264
=-1
Tính các tổng:
1/ S = 1 - 2 - 3 + 4 + 5 - 6 - 7 + 8 + ...+ 2001- 2002 - 2003 + 2004
2/ S = 1 + 2 - 3 - 4 + 5 + 6 - 7 - 8 + 9 + ...+ 2002 - 2003 - 2004 + 2005 + 2006
S=(1+2-3-4)+(5+6-7-8)+......+(2001+2002-2003-2004)+(2005+2006)
S=(-4)+(-4)+.......+(-4)+(2005+2006)
Dãy S có 2004-1:1+1=2004 số hạng
Dãy S có 2004:4=501 số -4
Do đó S=-4.501=-2004
S=-2004+(2005+2006)
S=-2004+4011
S=2007
1,S=(1-2-3+4)+(5-6-7+8)+.......+(2001-2002-2003+2004)
S=0+0+.........................+0
S=0
2,hình như pan gi sai đề
Bài 2. Tính:
a) A = 1 – 2 – 3 + 4 + 5 – 6 – 7 + 8 + ... + 2001 – 2002 – 2003 + 2004.
b) B = 1 + 2 – 3 – 4 + 5 + 6 – 7 – 8 + 9 + ... + 2002 – 2003 – 2004 + 2005 + 2006.Mik sẽ tick cho bạn trả lời nha