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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 \(\hept{\begin{cases}x-2010=4\\x-2011=4\\z-2012=4\end{cases}\Leftrightarrow\hept{\begin{cases}x=2014\\y=2015\\z=2016\end{cases}}}\)
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
a) ĐK: \(x>2009;y>2010;z>2011\)
\(\Leftrightarrow\frac{\sqrt{x-2009}-1}{x-2009}-\frac{1}{4}+\frac{\sqrt{y-2010}-1}{y-2010}-\frac{1}{4}+\frac{\sqrt{z-2011}-1}{z-2011}-\frac{1}{4}=0\)
\(\Leftrightarrow\frac{-\left(\sqrt{x-2009}-2\right)^2}{4\left(x-2009\right)}+\frac{-\left(\sqrt{y-2010}-2\right)^2}{4\left(y-2010\right)}+\frac{-\left(\sqrt{z-2011}-2\right)^2}{4\left(z-2011\right)}=0\left(1\right)\)
Dễ thấy với đkxđ thì \(VT\left(1\right)\le0\)
Dấu "=" xảy ra khi \(\hept{\begin{cases}\sqrt{x-2009}=2\\\sqrt{y-2010}=2\\\sqrt{z-2011}=2\end{cases}\Leftrightarrow\hept{\begin{cases}x=2013\\y=2014\\z=2015\end{cases}\left(tm\right)}}\)
\(\sqrt{x^2-9}+\sqrt{x^2-6x+9}=0\)(*)
\(ĐK:\orbr{\begin{cases}x\ge3\\x\le-3\end{cases}}\)
(*)\(\Leftrightarrow\sqrt{\left(x+3\right)\left(x-3\right)}+\sqrt{\left(x-3\right)^2}=0\)
\(\Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}+\sqrt{x-3}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\left(tm\right)\\\sqrt{x+3}+\sqrt{x-3}=0\end{cases}}\)
Xét phương trình\(\sqrt{x+3}+\sqrt{x-3}=0\)(**) có \(\sqrt{x+3}\ge0;\sqrt{x-3}\ge0\)nên (**) xảy ra khi \(\hept{\begin{cases}\sqrt{x+3}=0\\\sqrt{x-3}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-3\\x=3\end{cases}}\left(L\right)\)
Vậy phương trình có một nghiệm duy nhất là 3
\(ĐKXĐ:x\ne2009;y\ne2010;z\ne2011;x,y,z\in R\)
\(\frac{\sqrt{x-2009}-1}{x-2009}+\frac{\sqrt{y-2010}-1}{y-2010}+\frac{\sqrt{z-2011}-1}{z-2011}=\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{x-2009}-\frac{\sqrt{x-2009}}{x-2009}+\frac{1}{y-2010}-\frac{\sqrt{y-2011}}{y-2011}+\frac{1}{z-2011}-\frac{\sqrt{z-2011}}{z-2011}=\frac{-3}{4}\)
\(\Leftrightarrow\left(\frac{1}{\sqrt{x-2009}^2}-\frac{1}{\sqrt{x-2009}}+\frac{1}{4}\right)+\left(\frac{1}{\sqrt{y-2010}^2}-\frac{1}{\sqrt{y-2010}}+\frac{1}{4}\right)+\left(\frac{1}{\sqrt{z-2011}^2}+\frac{1}{\sqrt{z-2011}}+\frac{1}{4}\right)=0\)\(\Leftrightarrow\left(\frac{1}{\sqrt{x-2009}}-\frac{1}{2}\right)^{^2}+\left(\frac{1}{\sqrt{y-2010}}-\frac{1}{2}\right)^2+\left(\frac{1}{\sqrt{z-2011}}-\frac{1}{2}\right)^2=0\)
- \(\frac{1}{\sqrt{x-2009}}-\frac{1}{2}=0\)
- \(\frac{1}{\sqrt{y-2010}}-\frac{1}{2}=0\)
- \(\frac{1}{\sqrt{z-2011}}-\frac{1}{2}=0\)
\(\Leftrightarrow\frac{1}{\sqrt{x-2009}}=\frac{1}{2};\frac{1}{\sqrt{y-2010}}=\frac{1}{2};\frac{1}{\sqrt{z-2011}}=\frac{1}{2}\)
\(\Leftrightarrow x=2013;y=2014;z=2015\inĐKXĐ\)
VẬY \(x=2013;y=2014;z=2015\)
\(pt\Leftrightarrow\frac{1-\sqrt{x-2009}}{x-2009}+\frac{1-\sqrt{y-2010}}{y-2010}+\frac{1-\sqrt{z-2011}}{z-2011}=-\frac{3}{4}\)
\(\Leftrightarrow\left(\frac{1}{x-2009}-\frac{\sqrt{x-2009}}{x-2009}+\frac{1}{4}\right)+\left(\frac{1}{y-2010}-\frac{\sqrt{y-2010}}{y-2010}+\frac{1}{4}\right)+\left(\frac{1}{z-2011}-\frac{\sqrt{z-2011}}{z-2011}+\frac{1}{4}\right)=0\)
\(\Leftrightarrow\left(\frac{1}{x-2009}-\frac{1}{\sqrt{x-2009}}+\frac{1}{4}\right)+\left(\frac{1}{y-2010}-\frac{1}{\sqrt{y-2010}}+\frac{1}{4}\right)+\left(\frac{1}{z-2011}-\frac{1}{\sqrt{z-2011}}+\frac{1}{4}\right)=0\)
\(\Leftrightarrow\left(\frac{1}{\sqrt{x-2009}}-\frac{1}{2}\right)^2+\left(\frac{1}{\sqrt{y-2010}}-\frac{1}{2}\right)^2+\left(\frac{1}{\sqrt{z-2011}}-\frac{1}{2}\right)^2=0\)
Xảy ra khi \(\hept{\begin{cases}\frac{1}{\sqrt{x-2009}}=\frac{1}{2}\\\frac{1}{\sqrt{y-2010}}=\frac{1}{2}\\\frac{1}{\sqrt{z-2011}}=\frac{1}{2}\end{cases}}\Rightarrow\hept{\begin{cases}\sqrt{x-2009}=2\\\sqrt{y-2010}=2\\\sqrt{z-2011}=2\end{cases}}\Rightarrow\hept{\begin{cases}x=2013\\y=2014\\z=2015\end{cases}}\)
\(\left[\frac{\left(1-\sqrt{x}\right)\left(x+\sqrt{x}+1\right)}{1-\sqrt{x}}\right]\left[\frac{1-\sqrt{x}}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}\right]^2=\left(x+\sqrt{x}+1\right)\frac{1}{\left(1+\sqrt{x}\right)^2}=\frac{x+\sqrt{x}+1}{x+2\sqrt{x}+1}\)
Đề bài sai
\(\sqrt{2012}-\sqrt{2011}=\frac{1}{\sqrt{2012}+\sqrt{2011}}\)
\(\sqrt{2011}-\sqrt{2010}=\frac{1}{\sqrt{2011}+\sqrt{2010}}\)
Do \(\sqrt{2012}>\sqrt{2010}\) \(\Rightarrow\sqrt{2012}+\sqrt{2011}>\sqrt{2011}+\sqrt{2010}>0\)
\(\Rightarrow\frac{1}{\sqrt{2012}+\sqrt{2011}}< \frac{1}{\sqrt{2011}+\sqrt{2010}}\Rightarrow\sqrt{2012}-\sqrt{2011}< \sqrt{2011}-\sqrt{2010}\)
\(A=\frac{x+2\sqrt{xy}+y-4\sqrt{xy}}{\sqrt{x}-\sqrt{y}}+\frac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}=\frac{\left(\sqrt{x}-\sqrt{y}\right)^2}{\sqrt{x}-\sqrt{y}}+\frac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\)
\(=\sqrt{x}-\sqrt{y}+\sqrt{x}-\sqrt{y}=2\sqrt{x}-2\sqrt{y}\)
\(M^2=\left(\sqrt{x-1}+\sqrt{9-x}\right)^2\le2\left(x-1+9-x\right)=16\)
\(\Rightarrow M\le4\Rightarrow M_{max}=4\) khi \(x-1=9-x\Leftrightarrow x=5\)
\(A=\sqrt{x^2-2x+1}+\sqrt{x^2+4x+4}\)
\(=\sqrt{\left(x-1\right)^2}+\sqrt{\left(x+2\right)^2}\)
\(=|1-x|+|x+2|\ge|1-x+x+2|=3\)
\(x\sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}}=2\)
\(\Leftrightarrow x\sqrt{\left(\sqrt{x+\frac{1}{4}}+\frac{1}{2}\right)^2}=2\)
\(\Leftrightarrow x\sqrt{x+\frac{1}{4}}+\frac{1}{2}=2\)
\(\Leftrightarrow x\sqrt{x+\frac{1}{4}}=\frac{3}{2}\)
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