= 3 mỹ nữ chân dàj =)) hehe định dụ hả =)) cuốj tuần zz nghen.nghỉ hè chưa nhok

Áp dụng bất đẳng thức AM - GM, ta có:

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

\(\ge3\sqrt[3]{\dfrac{1}{\left(x-1\right)^2}\times\dfrac{1}{\left(2-x\right)^2}\times\dfrac{1}{\left(x-1\right)\left(2-x\right)}}\)

\(=\dfrac{3}{\left(x-1\right)\left(x-2\right)}=\dfrac{3}{-x^2+3x-2}\)

Vì \(-x^2+3x-2=-\left(x-\dfrac{3}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\)

nên \(S\ge\dfrac{3}{\dfrac{1}{4}}=12\)

Đẳng thức xảy ra khi \(\left\{{}\begin{matrix}\dfrac{1}{\left(x-1\right)^2}=\dfrac{1}{\left(2-x\right)^2}=\dfrac{1}{\left(x-1\right)\left(2-x\right)}\\x-\dfrac{3}{2}=0\end{matrix}\right.\)

\(\Leftrightarrow x=\dfrac{3}{2}\left(\text{ nhận }\right)\)

Vậy \(Min_S=12\Leftrightarrow x=\dfrac{3}{2}\)

Ta có: \(\left\{{}\begin{matrix}2014-x< 2015-x\\2014-y< 2015-y\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\sqrt{2014-x}< \sqrt{2015-x}\\\sqrt{2014-y}< \sqrt{2015-y}\end{matrix}\right.\)

\(\Rightarrow\sqrt{2014-x}+\sqrt{2014-y}< \sqrt{2015-x}+\sqrt{2015-y}\)

\(\Rightarrow\dfrac{1}{\sqrt{2014-x}+\sqrt{2014-y}}>\dfrac{1}{\sqrt{2015-x}+\sqrt{2015-y}}\)

\(\Rightarrow\dfrac{1}{\sqrt{2014-x}+\sqrt{2014-y}}-\dfrac{1}{\sqrt{2015-x}+\sqrt{2015-y}}>0\)