Tính nhanh:
\(1^2-2^2+3^2-4^2+.....2002^2+2003^2-2004^2+2005^2\)
(1/2003+1/2004-1/2005)/(5/2003+5/2004-5/2005)-(2/2002+2/2003-2/2004)/(3/2002+3/2003-3/2004)
Tính : P = \(\frac{\frac{1}{2003}+\frac{1}{2004}+\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}-\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{2004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)
Tính nhanh tổng đại số sau:
a) S=1-2-3+4+5-6-7+8+...+2001-2002-2003+2004
b) S=1+2-3-4+5+6-7-8+9+...+2002-2003-2004+2005+2006
a) Ta có: S = 1 - 2 - 3 + 4 + 5 - 6 - 7+ 8 + ... + 2001 - 2002 - 2003 + 2004
\(\Rightarrow\) S = (1 - 2 - 3 + 4) + (5 - 6 - 7+ 8) + ... + (2001 - 2002 - 2003 + 2004)
\(\Rightarrow\) S = (-4 + 4) + (-8 + 8) + ... + (-2004 + 2004)
\(\Rightarrow\) S = 0 + 0 + ... + 0
\(\Rightarrow\) S = 0
Tính :
\(P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}-\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{2004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)
\(P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}-\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{2004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)
\(=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{5\left(\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}\right)}-\frac{2\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}\right)}{3\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}\right)}\)
\(=\frac{1}{5}-\frac{2}{3}=-\frac{7}{15}\)
Ta có:
\(P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}-\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{2004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)
\(P=\frac{1}{5}\cdot\left(\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}\right)-\frac{2}{3}\cdot\left(\frac{\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}}{\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}}\right)\)
\(P=\frac{1}{5}-\frac{2}{3}=-\frac{7}{15}\)
Tính:
1^2-2^2+3^2-4^2+.......-2002^2+2003^2-2004^2+2005^2
Đặt dãy trên là A
Ta có:
A=(12-22)+(32-42)+...+(20032-20042)+20052
A=(1-2)(1+2)+(3-4)(3+4)+...+(2003-2004)(2003+2004)+20052
A=(-1.3)+(-1.7)+(-1.11)+...+(-1.4007)+4020025
A=-3+(-7)+(-11)+...+(-4007)+4020025
A=-(3+7+11+...+4007)+4020025
A=-{(4007+3)[(4007-3):4+1]}+4020025
A=-(4010.1002)+4020025
A=-4018020+4020025
A=2005
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
1, Tính : P = \(\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}-\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{2004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)
2,Biết : 13 + 23 + .......+103 = 3025
Tính S = 23 + 43 + 63 + ....+ 203
Bài 1:
\(P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}-\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{2004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)
\(\Rightarrow P=\frac{1\left(\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}\right)}{5\left(\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}\right)}-\frac{2\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2002}\right)}{3\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}\right)}\)
\(\Rightarrow P=\frac{1}{5}-\frac{2}{3}\)
\(\Rightarrow P=\frac{-7}{15}\)
Vậy \(P=\frac{-7}{15}\)
Bài 2:
Ta có: \(S=23+43+63+...+203\)
\(\Rightarrow S=13+10+20+23+...+103+100\)
\(\Rightarrow S=\left(13+23+...+103\right)+\left(10+20+...+100\right)\)
\(\Rightarrow S=3025+450\)
\(\Rightarrow S=3475\)
Vậy S = 3475
1. \(P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}-\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{2004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)
=> P =\(\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{5\left(\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}\right)}-\frac{2\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}\right)}{3\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}\right)}\)
=> P = \(\frac{1}{5}-\frac{2}{3}\)
P = \(\frac{3}{15}-\frac{10}{15}\)
=> P =\(\frac{-7}{15}\)
2. ta có:
S = 23 + 43 + 63 +...+ 203
=> S = 13 + 10 + 23 + 20 +...+ 103 + 100
=> S = ( 13 + 23+...+ 103 ) + ( 10 + 20 +...+ 100 )
=> S = 3025 + 550
=> S = 3575
Vậy S = 3575
1. \(\dfrac{\dfrac{1}{2003}+\dfrac{1}{2004}-\dfrac{1}{2005}}{\dfrac{5}{2003}+\dfrac{5}{2004}-\dfrac{5}{2005}}-\dfrac{\dfrac{2}{2003}+\dfrac{2}{2004}-\dfrac{2}{2005}}{\dfrac{3}{2003}+\dfrac{3}{2004}-\dfrac{3}{2005}}\)
=\(\dfrac{\dfrac{1}{2003}+\dfrac{1}{2004}-\dfrac{1}{2005}}{5\cdot\left(\dfrac{1}{2003}+\dfrac{1}{2004}-\dfrac{1}{2005}\right)}-\)\(\dfrac{2\cdot\left(\dfrac{1}{2003}+\dfrac{1}{2004}-\dfrac{1}{2005}\right)}{3\cdot\left(\dfrac{1}{2003}+\dfrac{1}{2004}-\dfrac{1}{2005}\right)}\)
=\(\dfrac{1}{5}-\dfrac{2}{3}\)
=\(-\dfrac{7}{15}\)
1-2-3+4+5-6-7+8+2011-2002-2003+2004+2005-2006-2007+2008+2009=
Tính nhanh:
=(1-2-3+4)+(5-6-7+8)+...+(2005-2006-2007+2008)+2009
=2009
tính : a)1-2-3+4+5-6-7+8+...+2001-2002-2003+2004
b)1+2-3-4+5+6-7-8+9+...+2002-2003-2004+2005+2006
a) 1-2-3+4+5-6-7+8+...+2001-2002-2003+2004
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 chữ số (-4)
do đó S = -4. 501 = -2004
S = -2004 + (2005+2006)
S = -2004 + 4011
S = 2007
b) tương tự nhé!!
675676587689689
a) Nhóm 4 số hạng liên tiếp từ đầu dãy:
A = (1-2-3+4)+(5-6-7+8)+(9-10-11+12)+ ...+(2001-2002-2003+2004) = 0
b) Nhóm 4 số hạng liên tiếp bắt đầu từ số thứ 2:
B = 1+(2-3-4+5)+(6-7-8+9)+...+(2002-2003-2004+2005)+2006 = 1+2006 = 2007.
chết cho mk xin lỗi mk làm câu b) mà kéo nhầm câu a đó bn!!
sorry nhìu!! 654647567689