Tìm x: a, \(\frac{x-2004}{2003}+\frac{x-2003}{2004}+\frac{x-2005}{2004}=3+\frac{2005}{2003}\)\(+\frac{2004}{2005}\)
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=\(\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}\)-\(\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{2004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)
\(P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}-\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{2004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)
\(P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{5\left(\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}\right)}-\frac{2\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}\right)}{3\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}\right)}\)
\(P=\frac{1}{5}-\frac{2}{3}=\frac{3-10}{15}=\frac{-7}{15}\)
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:
\(P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}-\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{2004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)
\(P=\frac{1}{5}\cdot\left(\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}\right)-\frac{2}{3}\cdot\left(\frac{\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}}{\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}}\right)\)
\(P=\frac{1}{5}-\frac{2}{3}=-\frac{7}{15}\)
Giải phương trình sau :
\(\frac{x^2-2008}{2007}+\:\frac{x^2-2007}{2006}+\frac{x^2-2006}{2005}=\:\frac{x^2-\:2005}{2004}+\:\frac{x^2-2004}{2003}+\:\frac{x^2-2003}{2002}\)
Ta có : \(\frac{x^2-2008}{2007}+\frac{x^2-2007}{2006}+\frac{x^2-2006}{2005}=\frac{x^2-2005}{2004}+\frac{x^2-2004}{2003}+\frac{x^2-2003}{2002}\)
=> \(\frac{x^2-2008}{2007}+1+\frac{x^2-2007}{2006}+1+\frac{x^2-2006}{2005}+1=\frac{x^2-2005}{2004}+1+\frac{x^2-2004}{2003}+1+\frac{x^2-2003}{2002}+1\)
=> \(\frac{x^2-2008}{2007}+\frac{2007}{2007}+\frac{x^2-2007}{2006}+\frac{2006}{2006}+\frac{x^2-2006}{2005}+\frac{2005}{2005}=\frac{x^2-2005}{2004}+\frac{2004}{2004}+\frac{x^2-2004}{2003}+\frac{2003}{2003}+\frac{x^2-2003}{2002}+\frac{2002}{2002}\)
=> \(\frac{x^2-1}{2007}+\frac{x^2-1}{2006}+\frac{x^2-1}{2005}=\frac{x^2-1}{2004}+\frac{x^2-1}{2003}+\frac{x^2-1}{2002}\)
=> \(\frac{x^2-1}{2007}+\frac{x^2-1}{2006}+\frac{x^2-1}{2005}-\frac{x^2-1}{2004}-\frac{x^2-1}{2003}-\frac{x^2-1}{2002}=0\)
=> \(\left(x^2-1\right)\left(\frac{1}{2007}+\frac{1}{2006}+\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}-\frac{1}{2002}\right)=0\)
=> \(x^2-1=0\)
=> \(x^2=1\)
=> \(x=\pm1\)
Vậy phương trình có 2 nghiệm là x = 1, x = -1 .
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. \(\dfrac{\dfrac{1}{2003}+\dfrac{1}{2004}-\dfrac{1}{2005}}{\dfrac{5}{2003}+\dfrac{5}{2004}-\dfrac{5}{2005}}-\dfrac{\dfrac{2}{2003}+\dfrac{2}{2004}-\dfrac{2}{2005}}{\dfrac{3}{2003}+\dfrac{3}{2004}-\dfrac{3}{2005}}\)
=\(\dfrac{\dfrac{1}{2003}+\dfrac{1}{2004}-\dfrac{1}{2005}}{5\cdot\left(\dfrac{1}{2003}+\dfrac{1}{2004}-\dfrac{1}{2005}\right)}-\)\(\dfrac{2\cdot\left(\dfrac{1}{2003}+\dfrac{1}{2004}-\dfrac{1}{2005}\right)}{3\cdot\left(\dfrac{1}{2003}+\dfrac{1}{2004}-\dfrac{1}{2005}\right)}\)
=\(\dfrac{1}{5}-\dfrac{2}{3}\)
=\(-\dfrac{7}{15}\)
So sánh:
\(A=\frac{2003}{2004}+\frac{2005}{2006};B=\frac{2003+2004}{2004+2005}\)
so sánh A và B
A = \(\frac{2003}{2004}+\frac{2004}{2005}\)và B = \(\frac{2003+2004}{2004+2005}\)
\(B=\frac{2003+2004}{2004+2005}=\frac{2003}{2004+2005}+\frac{2004}{2004+2005}\)
Ta có: \(\frac{2003}{2004}>\frac{2003}{2004+2005}\)
\(\frac{2004}{2005}>\frac{2004}{2004+2005}\)
\(\frac{2003}{2004}+\frac{2004}{2005}>\frac{2003+2004}{2004+2005}\)
\(A>B\)
Vậy A>B
\(\text{ Bài giải}\)
\(A=\frac{2003}{2004}+\frac{2004}{2005}=0,999500998 + 0,999501247=1.99900225\)
\(B=\frac{2003+2004}{2004+2005}=\frac{4007}{4009}=0,999501122\)
\(\text{Vì : }1,99900224>0,999501122\text{ nên }A>B\)
\(\text{Vậy : }A>B\)
giải giùm mình với:
So sánh A và B, biết
\(A=\frac{2003+2004}{2004+2005}\)
\(B=\frac{2003}{2004+2005}\)+\(\frac{2004}{2004+2005}\)
So sành \(\frac{2002}{2001}+\frac{2003}{2002}+\frac{2004}{2003}+\frac{2005}{2004}+\frac{2006}{2005}+\frac{2007}{2006}+\frac{2008}{2007}+\frac{2009}{2008}\)với 8
=1+1/2001+1+1/2002+1+1/2003+...+1+1/2008=8+1/2001+1/2002+1/2003+...+1/2008>8
\(\frac{2002}{2001}+\frac{2003}{2002}+\frac{2004}{2003}+\frac{2005}{2004}+\frac{2006}{2005}+\frac{2007}{2006}+\frac{2008}{2007}+\frac{2009}{2008}>8\)
Ta có:
2002/2001=1+1/2001
2003/2002=1+1/2002
2004/2003= 1+ 1/2003
2005/2004= 1+ 1/2004
2006/2005=1+ 1/2005
2007/2006= 1+ 1/2006
2008/2007=1 + 1/2007.
2009/2008=1+ 1/2008.
=> 2002/2001+2003/2002+2004?2003+2005/2004+2006/2005+ 2007/2006+ 2008/2007+ 2009/2008= 1+1+1+1+1+1+1+1+1/2001+1/2002+1/2003+1/2004+1/2005+1/2006+1/2007+1/2008>8.
Nhớ k đúng cho mình nha!! Thanks!!!