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Đặt: \(L_2=\dfrac{2007}{1}+\dfrac{2006}{2}+\dfrac{2005}{3}+...+\dfrac{2}{2006}+\dfrac{1}{2007}\)
\(L_2=1+\left(\dfrac{2006}{2}+1\right)+\left(\dfrac{2005}{3}+1\right)+...+\left(\dfrac{2}{2006}+1\right)+\left(\dfrac{1}{2007}+1\right)\)
\(L_2=\dfrac{2008}{2008}+\dfrac{2008}{2}+\dfrac{2008}{3}+...+\dfrac{2008}{2006}+\dfrac{2008}{2007}\)
\(L_2=2008\left(\dfrac{1}{2}+\dfrac{1}{3}+..+\dfrac{1}{2006}+\dfrac{1}{2007}+\dfrac{1}{2008}\right)\)
\(\dfrac{L_1}{L_2}=\dfrac{1}{2008}\)
\(B=1+\left(\frac{2007}{2}+1\right)+\left(\frac{2006}{3}+1\right)+...+\left(\frac{1}{2008}+1\right)=2009\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2008}+\frac{1}{2009}\right)\Rightarrow\frac{A}{B}=\frac{1}{2009}\)
\(B=\dfrac{2008}{1}+\dfrac{2007}{2}+\dfrac{2006}{3}+...+\dfrac{2}{2007}+\dfrac{1}{2008}\)
\(B=1+\left(\dfrac{2007}{2}+1\right)+\left(\dfrac{2006}{3}+1\right)+...+\left(\dfrac{2}{2007}+1\right)+\left(\dfrac{1}{2008}+1\right)\)
\(B=\dfrac{2009}{2009}+\dfrac{2009}{2}+\dfrac{2009}{3}+..+\dfrac{2009}{2007}+\dfrac{2009}{2008}\)
\(B=2009\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}\right)\)
\(\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}}{2009\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}\right)}\)
\(\dfrac{A}{B}=\dfrac{1}{2009}\)
số số hạng của A là :
( 2007 - 3 ) : 3 + 1 = 669 ( số )
tổng A là :
( 2007 + 3 ) . 669 : 2 = 672345
B = \(\dfrac{\dfrac{2006}{2}+\dfrac{2006}{3}+\dfrac{2006}{4}+...+\dfrac{2006}{2007}}{\dfrac{2006}{1}+\dfrac{2005}{2}+\dfrac{2004}{3}+...+\dfrac{1}{2006}}\)
B = \(\dfrac{2006.\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}\right)}{\left(\dfrac{2005}{2}+1\right)+\left(\dfrac{2004}{3}+1\right)+...+\left(\dfrac{1}{2006}+1\right)+1}\)
B = \(\dfrac{2006.\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}\right)}{\dfrac{2007}{2}+\dfrac{2007}{3}+...+\dfrac{2007}{2006}+\dfrac{2007}{2007}}\)
B = \(\dfrac{2006.\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}\right)}{2007.\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2006}+\dfrac{1}{2007}\right)}\)
B = \(\dfrac{2006}{2007}\)
2006/1 là 2006, tách 1 của 2006 ra 2005 phân số còn lại 1
1)\(\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}}{\dfrac{2008}{1}+\dfrac{2007}{2}+\dfrac{2006}{3}+...+\dfrac{2}{2007}+\dfrac{1}{2008}}\)
\(\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}}{2008+\dfrac{2007}{2}+\dfrac{2006}{3}+...+\dfrac{2}{2007}+\dfrac{1}{2008}}\)
\(\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}}{1+\left(\dfrac{2007}{2}+1\right)+\left(\dfrac{2006}{3}+1\right)+...+\left(\dfrac{2}{2007}+1\right)+\left(\dfrac{1}{2008}+1\right)}\)
\(\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}}{\dfrac{2009}{2009}+\dfrac{2009}{2}+\dfrac{2009}{3}+...+\dfrac{2009}{2007}+\dfrac{2009}{2008}}\)
\(\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}}{2009\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}\right)}\)
\(\dfrac{A}{B}=\dfrac{1}{2009}\)
2) \(A=\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}\)
\(A=\dfrac{2^2-1^2}{1^2.2^2}+\dfrac{3^2-2^2}{2^2.3^2}+\dfrac{4^2-3^2}{3^2.4^2}+...+\dfrac{10^2-9^2}{9^2.10^2}\)
\(A=1-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+\dfrac{1}{3^2}-\dfrac{1}{4^2}+...+\dfrac{1}{9^2}-\dfrac{1}{10^2}\)
\(A=1-\dfrac{1}{10^2}< 1\left(đpcm\right)\)
1,
x+1/2+x+1/3+x+1/4-x+1/5-x+1/6=0
(x+1)(1/2+1/3+1/4-1/5-1/6)=0
vì 1/2+1/3+1/4-1/5-1/6 khác 0
suy ra x+1=0 suy ra x=-1
a, \(A=\dfrac{10^{15}+1}{10^6+1}>1\);\(B=\dfrac{10^6+1}{10^{17}+1}< 1\)
⇒\(A>B\)
b, \(D=\dfrac{2^{2007}+3}{2^{2006}-1}=\dfrac{2^{2008}+6}{2^{2007}-2}\)
Ta có : \(\dfrac{2^{2008}-3}{2^{2007}-1}< \dfrac{2^{2008}-3}{2^{2007}-2}< \dfrac{2^{2008}+6}{2^{2007}-2}\)
⇒ \(C< D\)
c, \(M=\dfrac{3}{8^3}+\dfrac{7}{8^4}=\dfrac{3}{8^3}+\dfrac{3}{8^4}+\dfrac{4}{8^4}\)
\(N=\dfrac{7}{8^3}+\dfrac{3}{8^4}=\dfrac{3}{8^3}+\dfrac{4}{8^3}+\dfrac{3}{8^4}\)
Vì \(\dfrac{4}{8^4}< \dfrac{4}{8^3}\)
⇒ \(M< N\)
\(\dfrac{x-1}{2009}+\dfrac{x-2}{2008}=\dfrac{x-3}{2007}+\dfrac{x-4}{2006}\)
<=>\(\dfrac{x-1}{2009}-1+\dfrac{x-2}{2008}-1=\dfrac{x-3}{2007}-1+\dfrac{x-4}{2006}-1\)
<=>\(\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}=\dfrac{x-2010}{2007}+\dfrac{x-2010}{2006}\)
<=>\(\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}-\dfrac{x-2010}{2007}-\dfrac{x-2010}{2006}=0\)
<=>\(\left(x-2010\right)\left(\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\right)=0\)
Vì \(\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\ne0\) nên x-2010=0 <=>x=2010
a) Ta có : \(x-\dfrac{1}{2}=0\Rightarrow x=\dfrac{1}{2}\\ x+2=0\Rightarrow x=-2\)
Lập bảng xét dấu:
| x | -2 | \(\dfrac{1}{2}\) | |||
| x + 2 | - | 0 | + | + | |
| x - \(\dfrac{1}{2}\) | - | - | 0 | + |
TH : Xét x < -2
Ta có : - ( x+ 2) - (x - \(\dfrac{1}{2}\)) = \(\dfrac{3}{4}\)
-x - 2 -x + \(\dfrac{1}{2}\) = \(\dfrac{3}{4}\)
- 2x - 2 + \(\dfrac{1}{2}\)= \(\dfrac{3}{4}\)
-2x = 2\(\dfrac{1}{4}\)
=> x = \(-1\dfrac{1}{8}\) ( loại )
TH 2: \(-2\le x< \dfrac{1}{2}\)
Ta có : x + 2 + ( -x + \(\dfrac{1}{2}\)) = \(\dfrac{3}{4}\)
=> \(2,5=\dfrac{3}{4}\) ( loại )
TH3 : \(x\ge\dfrac{1}{2}\)
x+ 2 + x - \(\dfrac{1}{2}\) = \(\dfrac{3}{4}\)
2x + 1,5 = \(\dfrac{3}{4}\)
x = -0,375( loại )
vậy ....
b) \(\left(\dfrac{2}{3}-2x\right).1\dfrac{1}{2}=\dfrac{3}{4}\\ \Rightarrow\dfrac{2}{3}-2x=-\dfrac{3}{4}\\ \Rightarrow2x=1\dfrac{5}{12}\\ \Rightarrow x=\dfrac{17}{24}\)
c) \(\left|x-1\right|+2.\left(x+4\right)=10\\ \Rightarrow\left|x-1\right|=10-2x-8\\ \Rightarrow\left|x-1\right|=2-2x\)
TH1 : \(x-1\ge0\) \(\Rightarrow x\ge1\)
\(\Rightarrow x-1=2-2x\\ \Rightarrow3x=3\\ \Rightarrow x=1\left(TM\right)\)
TH2 : \(x-1< 0\Rightarrow x< 1\)
=> \(x-1=-2+2x\\ \Rightarrow-x=-1\Rightarrow x=1\)(loại)
Vậy x = 1
a, Theo bài ra ta có:
\(M=\dfrac{2007}{1}+1+\dfrac{2006}{2}+1+.......+\dfrac{2}{2006}+1+\dfrac{1}{2007}+1-2007\)
( Ta thêm 1 vào mỗi một số hạng trong M nên phải bớt đi 2017 vì có 2017 số hạng ) ;'
\(=>M=2008+\dfrac{2008}{2}+\dfrac{2008}{3}+......+\dfrac{2008}{2007}+\dfrac{2008}{2007}-2007\)
\(=>M=\dfrac{2008}{2}+\dfrac{2008}{3}+\dfrac{2008}{4}+.....+\dfrac{2008}{2006}+\dfrac{2008}{2007}+1\)
Ta thấy xuất hiện 2008 chung nên đặt ra ngoài ta có:
\(=>M=2008\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+....+\dfrac{1}{2006}+\dfrac{1}{2007}+\dfrac{1}{2008}\right)\)
\(=>M:N=2008\)
Câu b đợi 1 chút nha.......
b, \(M=\dfrac{1}{11.13}+\dfrac{1}{13.15}+...+\dfrac{1}{31.33}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{11.13}+\dfrac{2}{13.15}+...+\dfrac{2}{31.33}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+...+\dfrac{1}{31}-\dfrac{1}{33}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{11}-\dfrac{1}{33}\right)\)
\(=\dfrac{1}{33}\)
\(N=\dfrac{12}{11.13.15}+\dfrac{12}{13.15.17}+...+\dfrac{12}{31.33.35}\)
\(=3\left(\dfrac{4}{11.13.15}+\dfrac{4}{13.15.17}+...+\dfrac{4}{31.33.35}\right)\)
\(=3\left(\dfrac{1}{11.13}-\dfrac{1}{13.15}+\dfrac{1}{13.15}-\dfrac{1}{15.17}+...+\dfrac{1}{31.33}-\dfrac{1}{33.35}\right)\)
\(=3\left(\dfrac{1}{11.13}-\dfrac{1}{33.35}\right)\)
\(=\dfrac{92}{5005}\)
\(\Rightarrow M:N=\dfrac{1}{33}:\dfrac{92}{5005}=\dfrac{455}{276}\)
Vậy...