Đáp án:1) $Min A=\dfrac{1}{2}\Leftrightarrow x=y=z=\dfrac{2}{3}$
2) $Min M=-335\Leftrightarrow x=-3$
3) $Max B=3\Leftrightarrow x=0$
Giải thích các bước giải:
$\begin{array}{l}
1)A=\dfrac{{{{\left( {x - 1} \right)}^2}}}{z} + \dfrac{{{{\left( {y - 1} \right)}^2}}}{x} + \dfrac{{{{\left( {z - 1} \right)}^2}}}{y}\\
\ge \dfrac{{{{\left( {x - 1 + y - 1 + z - 1} \right)}^2}}}{{z + x + y}}\left( {Cauchy - Schwarz} \right)\\
= \dfrac{{{{\left( {x + y + z - 3} \right)}^2}}}{{x + y + z}} = \dfrac{{{{\left( {2 - 3} \right)}^2}}}{2} = \dfrac{1}{2}
\end{array}$
Dấu bằng xảy ra khi và chỉ khi:
$\begin{array}{l}
\left\{ \begin{array}{l}
\dfrac{{x - 1}}{z} = \dfrac{{y - 1}}{x} = \dfrac{{z - 1}}{y}\\
x + y + z = 2
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\dfrac{{x - 1}}{z} = \dfrac{{y - 1}}{x} = \dfrac{{z - 1}}{y} = \dfrac{{x + y + z - 3}}{{x + y + z}} = \dfrac{{ - 1}}{3}\\
x + y + z = 2
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
2\left( {x - 1} \right) + z = 0\\
2\left( {y - 1} \right) + x = 0\\
x + y + z = 2
\end{array} \right. \Leftrightarrow x = y = z = \dfrac{2}{3}
\end{array}$
Vậy $Min A=\dfrac{1}{2}\Leftrightarrow x=y=z=\dfrac{2}{3}$
$\begin{array}{l}
2)M = \dfrac{{2010x + 2680}}{{{x^2} + 1}}\\
\Rightarrow M + 335 = \dfrac{{2010x + 2680}}{{{x^2} + 1}} + 335\\
\Rightarrow M + 335 = \dfrac{{335{x^2} + 2010x + 3015}}{{{x^2} + 1}}\\
\Rightarrow M + 335 = \dfrac{{335\left( {{x^2} + 6x + 9} \right)}}{{{x^2} + 1}} = \dfrac{{335{{\left( {x + 3} \right)}^2}}}{{{x^2} + 1}} \ge 0\left( {{{\left( {x + 3} \right)}^2} \ge 0,\forall x} \right)\\
\Rightarrow MinM = - 335 \Leftrightarrow x = - 3
\end{array}$
Vậy $Min M=-335\Leftrightarrow x=-3$
$\begin{array}{l}
3)B = \dfrac{{3\left( {x + 1} \right)}}{{{x^3} + {x^2} + x + 1}}\\
= \dfrac{{3\left( {x + 1} \right)}}{{\left( {x + 1} \right)\left( {{x^2} + 1} \right)}}\\
= \dfrac{3}{{{x^2} + 1}} \le 3\left( {{x^2} \ge 0,\forall x} \right)\\
\Rightarrow MaxB = 3 \Leftrightarrow x = 0
\end{array}$
Vậy $Max B=3\Leftrightarrow x=0$
