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16 tháng 8 2017
\(\dfrac{123}{456}\cdot\left(\dfrac{2010}{2011}-\dfrac{2011}{2010}\right)-\left(\dfrac{2009}{2010}-\dfrac{1}{2011}\right):\dfrac{456}{123}\)
\(=\dfrac{123}{456}\cdot\left(\dfrac{2010}{2011}-\dfrac{2011}{2010}\right)-\left(\dfrac{2009}{2010}-\dfrac{1}{2011}\right)\cdot\dfrac{123}{456}\)
\(=\dfrac{123}{456}\left[\left(\dfrac{2010}{2011}-\dfrac{2011}{2010}\right)-\left(\dfrac{2009}{2010}-\dfrac{1}{2011}\right)\right]\)
\(=\dfrac{123}{456}\left(\dfrac{2010}{2011}-\dfrac{2011}{2010}-\dfrac{2009}{2010}+\dfrac{1}{2011}\right)\)
\(=\dfrac{123}{456}\left[\left(\dfrac{2010}{2011}+\dfrac{1}{2011}\right)-\left(\dfrac{2011}{2010}+\dfrac{2009}{2010}\right)\right]\)
\(=\dfrac{123}{456}\left(1-2\right)\)
\(=-\dfrac{123}{456}\)
3 tháng 5 2016
Ta có B=\(\frac{2009^{2010}-2}{2009^{2011}-2}\)<1
=>\(\frac{2009^{2010}-2}{2009^{2011}-2}\)<\(\frac{2009^{2010}-2+3}{2009^{2011}-2+3}\)=\(\frac{2009^{2010}+1}{2009^{2011}+1}\)(1)
Mà \(\frac{2009^{2010}+1}{2009^{2011}+1}\)<1
=> \(\frac{2009^{2010}+1}{2009^{2011}+1}\)<\(\frac{2009^{2010}+1+2008}{2009^{2011}+1+2008}\)=\(\frac{2009^{2010}+2009}{2009^{2011}+2009}\)=\(\frac{2009\cdot\left(2009^{2009}+1\right)}{2009\cdot\left(2009^{2010}+1\right)}\)=\(\frac{2009^{2009}+1}{2009^{2010}+1}\)=A(2)
Từ (1)và(2)=>B<\(\frac{2009^{2010}+1}{2009^{2011}+1}\)<A=>B<A hay A>B
CC
15 tháng 4 2018
a)\(\dfrac{x+1}{x^2+x+1}-\dfrac{x-1}{x^2-x+1}=\dfrac{3}{x\left(x^4+x^2+1\right)}\left(1\right)\)
ĐK:\(x\ne0\)
\(\left(1\right)\Leftrightarrow\dfrac{x^3+1-\left(x^3-1\right)}{\left(x^2+1+x\right)\left(x^2+1-x\right)}=\dfrac{3}{x\left(x^4+x^2+1\right)}\\ \Leftrightarrow\dfrac{2}{\left(x^2+1\right)^2-x^2}=\dfrac{3}{x\left(x^4+x^2+1\right)}\\ \Leftrightarrow\dfrac{2x-3}{x\left(x^4+x^2+1\right)}=0\Rightarrow2x-3=0\Leftrightarrow x=\dfrac{3}{2}\left(TM\right)\)
CC
15 tháng 4 2018
\(\dfrac{9-x}{2009}+\dfrac{11-x}{2011}=2\Leftrightarrow\left(\dfrac{9-x}{2009}-1\right)+\left(\dfrac{11-x}{2011}-1\right)=0\Leftrightarrow\dfrac{-2000-x}{2009}+\dfrac{-2000-x}{2011}=0\\ \Leftrightarrow\left(-2000-x\right)\left(\dfrac{1}{2009}+\dfrac{1}{2011}\right)=0\Rightarrow x=-2000\)
4 tháng 9 2017
Ta có :
\(-2010\le x< 2010\)
\(\Leftrightarrow x\in\left\{-2010;2009;..........;2008;2009\right\}\)
Tổng các số nguyên x là :
\(\left(-2010\right)+\left(-2009\right)+..........+2009\)
\(=\left[\left(-2009\right)+2009\right]+\left[\left(-2008\right)+2008\right]+..........+\left(-2010\right)\)
\(=0+0+........+\left(-2010\right)\)
\(=-2010\)
4 tháng 9 2017
Ta có:\(\)-2010 ≤x<2010
=>x thuộc -2010;-2009;...2009
Do đó tổng của các số nguyên x đó là
\(\left(-2010\right)+\left(-2009\right)+...+2008+2009=-2010\)
Vậy...
CC
3 tháng 7 2018
ĐK : x>0
Đặt \(\sqrt{2010+\sqrt{x}}=t\left(t>0\right)\Rightarrow t^2=2010+\sqrt{x}\)
\(Pt\Rightarrow x+\sqrt{x}=t^2+t\)
Xét hàm số \(f\left(a\right)=a^2+a\) là hàm đồng biến \(\forall a>0\)
\(f\left(\sqrt{x}\right)=f\left(t\right)\Rightarrow x=t^2\Leftrightarrow x-\sqrt{x}-2010=0\\ \Leftrightarrow x=\left(\dfrac{1+\sqrt{8041}}{2}\right)^2\)
BT
1
4 tháng 1 2018
Đặt \(\left\{{}\begin{matrix}x-2008=n\\2x+2009=h\\3x-2011=t\end{matrix}\right.\Rightarrow n+h+t=6x-2010\)
\(\Rightarrow pt\Leftrightarrow\dfrac{1}{n}+\dfrac{1}{h}=\dfrac{1}{n+h+t}-\dfrac{1}{t}\)
\(\Leftrightarrow\dfrac{n+h}{hn}=\dfrac{-\left(n+h\right)}{t\left(n+h+t\right)}\)
\(\Leftrightarrow\left(n+h\right)\left(\dfrac{1}{hn}+\dfrac{1}{t\left(n+h+t\right)}\right)=0\)
\(\Leftrightarrow\left(n+h\right)\dfrac{t\left(n+h+t\right)+hn}{hnt\left(n+h+t\right)}=0\)
\(\Leftrightarrow\dfrac{\left(n+h\right)\left(n+t\right)\left(t+h\right)}{hnt\left(n+h+t\right)}=0\)
\(\Rightarrow\left[{}\begin{matrix}n=-h\\n=-t\\t=-h\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x-2008=-\left(2x+2009\right)\\x-2008=-\left(3x-2011\right)\\3x-2011=-\left(2x+2009\right)\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=\dfrac{4019}{4}\\x=\dfrac{2}{5}\end{matrix}\right.\)
\(A=\dfrac{2010}{2}+\dfrac{2010}{6}+...+\dfrac{2010}{9900}\\ =2010\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{9900}\right)\\ =2010\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)\\ =2010\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\\ =2010.\dfrac{99}{100}=\dfrac{19899}{10}\)