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NV
18 tháng 10 2019

Đặt \(x^{670}=a\ge0\)

\(a^3-2011a+\sqrt{2010}=0\)

\(\Leftrightarrow\left(a-\sqrt{2010}\right)\left(a^2+\sqrt{2010}a-1\right)=0\)

Bạn tự giải tiếp

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:\(x\ne0\)

\(\left(1\right)\Leftrightarrow\dfrac{x^3+1-\left(x^3-1\right)}{\left(x^2+1+x\right)\left(x^2+1-x\right)}=\dfrac{3}{x\left(x^4+x^2+1\right)}\\ \Leftrightarrow\dfrac{2}{\left(x^2+1\right)^2-x^2}=\dfrac{3}{x\left(x^4+x^2+1\right)}\\ \Leftrightarrow\dfrac{2x-3}{x\left(x^4+x^2+1\right)}=0\Rightarrow2x-3=0\Leftrightarrow x=\dfrac{3}{2}\left(TM\right)\)

15 tháng 4 2018

\(\dfrac{9-x}{2009}+\dfrac{11-x}{2011}=2\Leftrightarrow\left(\dfrac{9-x}{2009}-1\right)+\left(\dfrac{11-x}{2011}-1\right)=0\Leftrightarrow\dfrac{-2000-x}{2009}+\dfrac{-2000-x}{2011}=0\\ \Leftrightarrow\left(-2000-x\right)\left(\dfrac{1}{2009}+\dfrac{1}{2011}\right)=0\Rightarrow x=-2000\)

AH
Akai Haruma
Giáo viên
15 tháng 7 2017

Lời giải:

Áp dụng BĐT AM-GM:

\(\sqrt{x-2}=\sqrt{(x-2).1}\leq \frac{x-2+1}{2}\)

\(\sqrt{y+2009}=\sqrt{(y+2009).1}\leq \frac{y+2009+1}{2}\)

\(\sqrt{z-2010}=\sqrt{(z-2010).1}\leq \frac{z-2010+1}{2}\)

Cộng theo vế suy ra :

\(\sqrt{x-2}+\sqrt{y+2009}+\sqrt{z-2010}\leq \frac{x+y+z}{2}\)

Dấu bằng xảy ra khi \(x-2=y+2009=z-2010=1\Leftrightarrow \left\{\begin{matrix} x=3\\ y=-2008\\ z=2011\end{matrix}\right.\)

2 tháng 5 2016

A<B

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

 

 

4 tháng 1 2018

Đặt \(\left\{{}\begin{matrix}x-2008=n\\2x+2009=h\\3x-2011=t\end{matrix}\right.\Rightarrow n+h+t=6x-2010\)

\(\Rightarrow pt\Leftrightarrow\dfrac{1}{n}+\dfrac{1}{h}=\dfrac{1}{n+h+t}-\dfrac{1}{t}\)

\(\Leftrightarrow\dfrac{n+h}{hn}=\dfrac{-\left(n+h\right)}{t\left(n+h+t\right)}\)

\(\Leftrightarrow\left(n+h\right)\left(\dfrac{1}{hn}+\dfrac{1}{t\left(n+h+t\right)}\right)=0\)

\(\Leftrightarrow\left(n+h\right)\dfrac{t\left(n+h+t\right)+hn}{hnt\left(n+h+t\right)}=0\)

\(\Leftrightarrow\dfrac{\left(n+h\right)\left(n+t\right)\left(t+h\right)}{hnt\left(n+h+t\right)}=0\)

\(\Rightarrow\left[{}\begin{matrix}n=-h\\n=-t\\t=-h\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x-2008=-\left(2x+2009\right)\\x-2008=-\left(3x-2011\right)\\3x-2011=-\left(2x+2009\right)\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=\dfrac{4019}{4}\\x=\dfrac{2}{5}\end{matrix}\right.\)

14 tháng 8 2017

\(x^2+y^2+z^2=xy+yz+zx\)\(x^{2009}+y^{2009}+z^{2009}=3^{2010}\)

Ta có:

\(x^2+y^2+z^2=xy+yz+zx\)

\(\Leftrightarrow x^2+y^2+z^2-xy-yz-zx=0\)

\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2zx=0\)

\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2zx+x^2\right)=0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)

\(\left\{{}\begin{matrix}\left(x-y\right)^2\ge0\\\left(y-z\right)^2\ge0\\\left(z-x\right)^2\ge0\end{matrix}\right.\) \(\Rightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)

Dấu " = " xảy ra :

\(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y-z=0\\z-x=0\end{matrix}\right.\) \(\Rightarrow x=y=z\)

Thay \(x=y=z\) vào \(x^{2009}+y^{2009}+z^{2009}=3^{2009}\) ta được:

\(3x^{2009}=3x^{2010}\)

\(\Rightarrow x^{2009}=3^{2009}\)

\(\Rightarrow x=3\)

\(\Rightarrow y=z=x=3\)

Vậy \(\left(x;y;z\right)=\left(3;3;3\right)\)

14 tháng 8 2017

Thiếu đề chăng.?