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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
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/ \(\frac{1}{4}-\frac{\sqrt{x-2009}-1}{x-2009}+\frac{1}{4}-\frac{\sqrt{y-2010}-1}{y-2010}+\frac{1}{4}-\frac{\sqrt{z-2011}-1}{z-2011}=0\)
\(\Leftrightarrow\frac{x-2009-4\sqrt{x-2009}+4}{4\left(x-2009\right)}+\frac{y-2010-4\sqrt{y-2010}+4}{4\left(y-2010\right)}+\frac{z-2011-4\sqrt{z-2011}+4}{4\left(z-2011\right)}=0\)
\(\Leftrightarrow\frac{\left(\sqrt{x-2009}-2\right)^2}{4\left(x-2009\right)}+\frac{\left(\sqrt{y-2010}-2\right)^2}{4\left(y-2010\right)}+\frac{\left(\sqrt{z-2011}-2\right)^2}{4\left(z-2011\right)}=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2009}-2=0\\\sqrt{y-2010}-2=0\\\sqrt{z-2011}-2=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=2013\\y=2014\\z=2015\end{matrix}\right.\)
Đặt \(\left(x;y\right)=\left(a^2;b^2\right)\Rightarrow a^2+b^2=2010\)
\(P=\frac{a^2}{b}+\frac{b^2}{a}=\frac{a^4}{a^2b}+\frac{b^4}{ab^2}\ge\frac{\left(a^2+b^2\right)^2}{ab\left(a+b\right)}\ge\frac{\left(a^2+b^2\right)^2}{\left(\frac{a^2+b^2}{2}\right)\sqrt{2\left(a^2+b^2\right)}}=2\sqrt{1005}\)
Dấu "=" xảy ra khi \(x=y=1005\)