gt pt nó thành nhân tử thay vào P tính

mk nhớ lm bài tương tự thế này r` bn chịu khó mở ra xem lại ở đây olm.vn/?g=page.display.showtrack&id=424601&limit=260, ấn vào chữ Trang tiếp theo để tìm thêm nhé

thôi mk lm luôn bn chờ tí

Áp dụng BĐT Cô - si ngược dấu :

\(\sqrt{x-2010}=\frac{1}{2}\sqrt{4\left(x-2010\right)}\le\frac{4+\left(x-2010\right)}{4}\)

\(\Rightarrow\sqrt{x-2010}-1\le\frac{4+\left(x-2010\right)}{4}-1=\frac{x-2010}{4}\)

\(\Rightarrow\frac{\sqrt{x-2010}-1}{x-2010}\le\frac{1}{4}\)

Hoàn toàn tương tự với những phân thức còn lại 

\(\Rightarrow\frac{\sqrt{x-2010}-1}{x-2010}+\frac{\sqrt{y-2011}-1}{y-2011}\le\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=\frac{3}{4}\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}x-2010=4\\x-2011=4\\z-2012=4\end{cases}\Leftrightarrow\hept{\begin{cases}x=2014\\y=2015\\z=2016\end{cases}}}\)

Từ  \(\frac{1}{x}+\frac{1}{y}=\frac{1}{2010}\)

\(\Leftrightarrow\)  \(\frac{x+y}{xy}=\frac{1}{2010}\)

\(\Leftrightarrow2010x-xy+2010y-2010^2=-2010^2\)

\(\Leftrightarrow x\left(2010-y\right)+2010\left(y-2010\right)=-2010^2\)

\(\Leftrightarrow\left(x-2010\right)\left(y-2010\right)=2010^2\)

Ta có  \(\left(\sqrt{x-2010}+\sqrt{y-2010}\right)^2\)

\(=\left(x-2010\right)+\left(y-2010\right)+2\sqrt{\left(x-2010\right)\left(y-2010\right)}\)

\(=x+y-2.2010+2\sqrt{2010^2}=x+y\)

Do đó  \(x+y=\left(\sqrt{x-2010}+\sqrt{y-2010}\right)^2\) 

mà x, y > 0 nên  \(\sqrt{x+y}=\sqrt{x-2010}+\sqrt{y-2010}\)

Bạn thay y xyz=2010 vào A ta được

A= xyz*x/xy+xyz*x+xyz + y/yz+y+xyz + z/xz+z+1

suy ra A=x^2yz/xy(1+xz+z) + y/y(z+1+xz) + z/xz+x+1

 A= xz/1+xz+z + 1/z+1+xz + x/xz+z+1 = xz+1+x/xz+1+x =1

Vay A=1

ta có:

\(x\left(\sqrt{2011}+\sqrt{2010}\right)+y\left(\sqrt{2011}-\sqrt{2010}\right)=x\sqrt{2011}+x\sqrt{2010}+y\sqrt{2011}-y\sqrt{2010}\)

 pt tương đương với:

\(\left(x+y\right)\sqrt{2011}+\left(x-y\right)\sqrt{2010}=\sqrt{2011^3}+\sqrt{2010^3}\)

vì x,y là số hữu tỉ nên

\(\hept{\begin{cases}\sqrt{2011}\left(x+y\right)=\sqrt{2011^3}\\\sqrt{2010}\left(x-y\right)=\sqrt{2010^3}\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x+y=2011\\x-y=2010\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=\frac{4021}{2}\\y=\frac{1}{2}\end{cases}}\)   

tích trước trả lời sau

là sao bạn???

ĐKXĐ: ...

a/ \(A=x-2009-4\sqrt{x-2009}+4=\left(\sqrt{x-2009}-2\right)^2\ge0\)

\(A_{min}=0\) khi \(\sqrt{x-2009}-2=0\Rightarrow x=2013\)

b/ \(\frac{1}{4}-\frac{\sqrt{x-2009}-1}{x-2009}+\frac{1}{4}-\frac{\sqrt{y-2010}-1}{y-2010}+\frac{1}{4}-\frac{\sqrt{z-2011}-1}{z-2011}=0\)

\(\Leftrightarrow\frac{x-2009-4\sqrt{x-2009}+4}{4\left(x-2009\right)}+\frac{y-2010-4\sqrt{y-2010}+4}{4\left(y-2010\right)}+\frac{z-2011-4\sqrt{z-2011}+4}{4\left(z-2011\right)}=0\)

\(\Leftrightarrow\frac{\left(\sqrt{x-2009}-2\right)^2}{4\left(x-2009\right)}+\frac{\left(\sqrt{y-2010}-2\right)^2}{4\left(y-2010\right)}+\frac{\left(\sqrt{z-2011}-2\right)^2}{4\left(z-2011\right)}=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2009}-2=0\\\sqrt{y-2010}-2=0\\\sqrt{z-2011}-2=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=2013\\y=2014\\z=2015\end{matrix}\right.\)

\(Q=\Sigma\frac{x^4}{x^2+\sqrt{xy.zx}}\ge\frac{\left(x^2+y^2+z^2\right)^2}{x^2+y^2+z^2+xy+yz+zx}\ge\frac{x^2+y^2+z^2}{2}\ge\frac{\left(x+y+z\right)^2}{6}=\frac{3}{2}\)

Dấu "=" xảy ra khi x=y=z=1 

E hổng biết cách này có đúng ko nữa:((

5

Ta có:\(S=\frac{2010}{x}+\frac{1}{2010y}+\frac{1010}{1005}\ge2\sqrt{\frac{2010}{x}\cdot\frac{1}{2010y}}+\frac{1010}{1005}\left(AM-GM\right)\)

\(=\frac{2}{\sqrt{xy}}+\frac{2010}{1005}\ge\frac{2}{\frac{x+y}{2}}+2=4\)( AM-GM ngược dấu )

Dấu "=" xảy ra khi \(x=y=\frac{2010}{4024}\)

\(pt\Leftrightarrow\frac{1-\sqrt{x-2009}}{x-2009}+\frac{1-\sqrt{y-2010}}{y-2010}+\frac{1-\sqrt{z-2011}}{z-2011}=-\frac{3}{4}\)

\(\Leftrightarrow\left(\frac{1}{x-2009}-\frac{\sqrt{x-2009}}{x-2009}+\frac{1}{4}\right)+\left(\frac{1}{y-2010}-\frac{\sqrt{y-2010}}{y-2010}+\frac{1}{4}\right)+\left(\frac{1}{z-2011}-\frac{\sqrt{z-2011}}{z-2011}+\frac{1}{4}\right)=0\)

\(\Leftrightarrow\left(\frac{1}{x-2009}-\frac{1}{\sqrt{x-2009}}+\frac{1}{4}\right)+\left(\frac{1}{y-2010}-\frac{1}{\sqrt{y-2010}}+\frac{1}{4}\right)+\left(\frac{1}{z-2011}-\frac{1}{\sqrt{z-2011}}+\frac{1}{4}\right)=0\)

\(\Leftrightarrow\left(\frac{1}{\sqrt{x-2009}}-\frac{1}{2}\right)^2+\left(\frac{1}{\sqrt{y-2010}}-\frac{1}{2}\right)^2+\left(\frac{1}{\sqrt{z-2011}}-\frac{1}{2}\right)^2=0\)

Xảy ra khi \(\hept{\begin{cases}\frac{1}{\sqrt{x-2009}}=\frac{1}{2}\\\frac{1}{\sqrt{y-2010}}=\frac{1}{2}\\\frac{1}{\sqrt{z-2011}}=\frac{1}{2}\end{cases}}\Rightarrow\hept{\begin{cases}\sqrt{x-2009}=2\\\sqrt{y-2010}=2\\\sqrt{z-2011}=2\end{cases}}\Rightarrow\hept{\begin{cases}x=2013\\y=2014\\z=2015\end{cases}}\)