Đề bạn sai câu b/

thế c lm hộ t câu a vs

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

\(A=\sqrt{x^2-2x+1}+\sqrt{x^2+4x+4}\)

\(=\sqrt{\left(x-1\right)^2}+\sqrt{\left(x+2\right)^2}\)

\(=|1-x|+|x+2|\ge|1-x+x+2|=3\)

\(x\sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}}=2\)

\(\Leftrightarrow x\sqrt{\left(\sqrt{x+\frac{1}{4}}+\frac{1}{2}\right)^2}=2\)

\(\Leftrightarrow x\sqrt{x+\frac{1}{4}}+\frac{1}{2}=2\)

\(\Leftrightarrow x\sqrt{x+\frac{1}{4}}=\frac{3}{2}\)

Làm nốt

Thưa bn mk đã làm ra nhưng không biết có đúng không. Xem nhá:

Ta có:

\(\frac{\sqrt{x-2009}-1}{x-2009}+\frac{\sqrt{y-2001}-1}{y-2001}+\frac{\sqrt{z-2011}-1}{z-2011}=\frac{3}{4}\Leftrightarrow"\frac{1}{\sqrt{x-2009}}-\frac{1}{2}"^2+\)

\("\frac{1}{\sqrt{y-2010}}-\frac{1}{2}"^2-"\frac{1}{\sqrt{z-2011}}-\frac{1}{2}"^2=0\)

\(\Rightarrow x=2013;y=2014;z=2015\)

P/s: Bn thay Ngoặc Kép thành Ngoặc Đơn nhé

b) đk: \(x>2012;y>2013\)

pt \(\frac{16}{\sqrt{x-2012}}+\sqrt{x-2012}+\frac{1}{\sqrt{y-2013}}+\sqrt{y-2013}=10\)

\(VT\ge2\sqrt{\frac{16}{\sqrt{x-2012}}.\sqrt{x-2012}}+2\sqrt{\frac{1}{\sqrt{y-2013}}.\sqrt{y-2013}}=8+2=10\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(\hept{\begin{cases}x-2012=16\\y-2013=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2028\\y=2014\end{cases}}\)

\(ĐKXĐ:x\ne2009;y\ne2010;z\ne2011;x,y,z\in R\)

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

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

\(\Leftrightarrow\left(\frac{1}{\sqrt{x-2009}^2}-\frac{1}{\sqrt{x-2009}}+\frac{1}{4}\right)+\left(\frac{1}{\sqrt{y-2010}^2}-\frac{1}{\sqrt{y-2010}}+\frac{1}{4}\right)+\left(\frac{1}{\sqrt{z-2011}^2}+\frac{1}{\sqrt{z-2011}}+\frac{1}{4}\right)=0\)\(\Leftrightarrow\left(\frac{1}{\sqrt{x-2009}}-\frac{1}{2}\right)^{^2}+\left(\frac{1}{\sqrt{y-2010}}-\frac{1}{2}\right)^2+\left(\frac{1}{\sqrt{z-2011}}-\frac{1}{2}\right)^2=0\)

• \(\frac{1}{\sqrt{x-2009}}-\frac{1}{2}=0\)

• \(\frac{1}{\sqrt{y-2010}}-\frac{1}{2}=0\)
• \(\frac{1}{\sqrt{z-2011}}-\frac{1}{2}=0\)

\(\Leftrightarrow\frac{1}{\sqrt{x-2009}}=\frac{1}{2};\frac{1}{\sqrt{y-2010}}=\frac{1}{2};\frac{1}{\sqrt{z-2011}}=\frac{1}{2}\)

\(\Leftrightarrow x=2013;y=2014;z=2015\inĐKXĐ\)

  VẬY       \(x=2013;y=2014;z=2015\)

ko biet E=MC'2

\(x-2008=X;y-2009=Y;z-2010=Z\)

\(\sqrt{X}+\sqrt{Y}+\sqrt{Z}+3012=\frac{1}{2}\left(X+Y+Z+2008+2009+2010\right)\)

\(2.\sqrt{X}+2\sqrt{Y}+2\sqrt{Z}+2.3012=X+Y+Z+2009\cdot3\)

\(\left(X-2\sqrt{X}+1\right)+\left(Y-2\sqrt{Y}+1\right)+\left(Z-2\sqrt{Z}+1\right)+3.2008=2.3012\)

\(\left(\sqrt{X}-1\right)^2+\left(\sqrt{Y}-1\right)^2+\left(\sqrt{Z}-1\right)^2=2.3012-3.2008=0\)

\(X=1;Y=1;Z=1\Rightarrow x=2009;y=2010;z=2011\)