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Ta có :\(\frac{2012}{\sqrt{2013}}+\frac{2013}{\sqrt{2012}}=\frac{2013-1}{\sqrt{2013}}+\frac{2012+1}{\sqrt{2012}}\)
=>\(\frac{2013}{\sqrt{2013}}-\frac{1}{\sqrt{2013}}+\frac{2012}{\sqrt{2012}}+\frac{1}{\sqrt{2012}}\)
=>\(\sqrt{2013}-\frac{1}{\sqrt{2013}}+\sqrt{2012}+\frac{1}{\sqrt{2012}}\)
Mà \(\frac{1}{\sqrt{2012}}-\frac{1}{\sqrt{2013}}>0\)
Vậy \(\sqrt{2012}+\sqrt{2013}+\frac{1}{\sqrt{2012}}-\frac{1}{\sqrt{2013}}>\sqrt{2012}+\sqrt{2013}\)
Hay \(\frac{2012}{\sqrt{2013}}+\frac{2013}{\sqrt{2012}}>\sqrt{2012}+\sqrt{2013}\)
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 \(\Leftrightarrow\)\(\hept{\begin{cases}x-2012=16\\y-2013=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2028\\y=2014\end{cases}}\)
\(DK:\hept{\begin{cases}x>0&y>\frac{2012}{2013}&\end{cases}}\)
HPT
\(\text{ }\Leftrightarrow\hept{\begin{cases}2013\sqrt{2013y-2012}=\frac{2013}{x}\left(1\right)\\y^2+2012=\frac{2013}{x}\left(2\right)\end{cases}}\)
\(\left(1\right),\left(2\right)\Rightarrow y^2-2013\sqrt{2013y-2012}+2012=0\)
\(\Leftrightarrow\left(y^2-1\right)-2013\left(\sqrt{2013y-2012}-1\right)=0\)
\(\Leftrightarrow\left(y+1\right)\left(y-1\right)-\frac{2013^2\left(y-1\right)}{\sqrt{2013y-2012}+1}=0\)
\(\Leftrightarrow\left(y-1\right)\left(y+1-\frac{2013^2}{\sqrt{2013y-2012}+1}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}y=1\\y+1-\frac{2013^2}{\sqrt{2013y-2012}+1}=0\end{cases}}\)
Cai PT thu to ay vo nghiem nhung biet chung minh :)
\(\Rightarrow x=1\)
Vay nghiem cua HPT la \(\left(x;y\right)=\left(1;1\right)\)
b) \(\left(\sqrt{2x+3}-3\right)+\left(\sqrt{x+1}-2\right)+5=3x+2\left(\sqrt{2x^2+5x+3}-6\right)+12-16\)
\(\Leftrightarrow\left(\sqrt{2x+3}-3\right)+\left(\sqrt{x+1}-2\right)=3\left(x-3\right)+2\left(\sqrt{2x^2+5x+3}-6\right)\)
\(\Leftrightarrow\frac{2\left(x-3\right)}{\sqrt{2x+3}+3}+\frac{x-3}{\sqrt{x+1}+2}-3\left(x-3\right)-\frac{2\left(x-3\right)\left(2x+11\right)}{\sqrt{2x^2+5x+3}+6}=0\Leftrightarrow x-3=0\Leftrightarrow x=3.\)
Lời giải:
$A=|1-\sqrt{2012}|\sqrt{2012+2\sqrt{2012}+1}$
$=|1-\sqrt{2012}|\sqrt{(\sqrt{2012}+1)^2}$
$=|1-\sqrt{2012}|.|\sqrt{2012}+1|$
$=|(1-\sqrt{2012})(1+\sqrt{2012})|=|1-2012|=2011$
Ta có \(\frac{2012^{2013}}{2013^{2013}}=\frac{2012^{2012}}{2013^{2012}}.\frac{2012}{2013}\)
Vì \(\frac{2012}{2013}< 1\)nên\(\frac{2012^{2012}}{2013^{2012}}.\frac{2012}{2013}< \frac{2012^{2012}}{2013^{2012}}.1=\frac{2012^{2012}}{2013^{2012}}\)
hay \(\frac{2012^{2013}}{2013^{2013}}< \frac{2012^{2012}}{2013^{2012}}\)
\(\Rightarrow\frac{2012^{2013}}{2013^{2013}}+1< \frac{2012^{2012}}{2013^{2012}}+1\)
\(\Rightarrow\left(\frac{2012^{2013}}{2013^{2013}}+1\right)^{2012}< \left(\frac{2012^{2012}}{2013^{2012}}+1\right)^{2013}\)