Tìm Min
D=2x^2+9y^2-6xy-6x+12y+2012
E=x^2-2xy+4y^2-2x+10y+29
F=3/(2x-x^2-4)
G=2/(6x-5-9x^2)
Tìm Min D=2x^2+9y^2-6xy-6x+12y+2012 E=x^2-2xy+4y^2-2x+10y+29 F=3/(2x-x^2-4) G=2/(6x-5-9x^2)
By Abigail
By Abigail
Tìm Min
D=2x^2+9y^2-6xy-6x+12y+2012
E=x^2-2xy+4y^2-2x+10y+29
F=3/(2x-x^2-4)
G=2/(6x-5-9x^2)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
*)\\
D = 2{x^2} + 9{y^2} – 6xy – 6x + 12y + 2012\\
= \left( {{x^2} – 6xy + 9{y^2}} \right) – 4.\left( {x – 3y} \right) + \left( {{x^2} – 2x + 1} \right) + 2011\\
= {\left( {x – 3y} \right)^2} – 4.\left( {x – 3y} \right) + {\left( {x – 1} \right)^2} + 2011\\
= \left[ {{{\left( {x – 3y} \right)}^2} – 4.\left( {x – 3y} \right) + 4} \right] + {\left( {x – 1} \right)^2} + 2007\\
= {\left( {x – 3y – 2} \right)^2} + {\left( {x – 1} \right)^2} + 2007 \ge 2007,\,\,\forall x,y\\
\Rightarrow {D_{\min }} = 2007 \Leftrightarrow \left\{ \begin{array}{l}
{\left( {x – 3y – 2} \right)^2} = 0\\
{\left( {x – 1} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 1\\
y = – \frac{1}{3}
\end{array} \right.\\
*)\\
E = {x^2} – 2xy + 4{y^2} – 2x + 10y + 29\\
= \left( {{x^2} – 2xy + {y^2}} \right) – 2.\left( {x – y} \right) + \left( {3{y^2} + 8y + \frac{{16}}{3}} \right) + \frac{{71}}{3}\\
= {\left( {x – y} \right)^2} – 2\left( {x – y} \right) + 1 + 3.\left( {{y^2} + \frac{8}{3}y + \frac{{16}}{9}} \right) + \frac{{68}}{3}\\
= {\left( {x – y} \right)^2} – 2.\left( {x – y} \right) + 1 + 3.{\left( {y + \frac{4}{3}} \right)^2} + \frac{{68}}{3}\\
= {\left( {x – y – 1} \right)^2} + 3.{\left( {y + \frac{4}{3}} \right)^2} + \frac{{68}}{3} \ge \frac{{68}}{3},\,\,\,\forall x,y\\
\Rightarrow {E_{\min }} = \frac{{68}}{3} \Leftrightarrow \left\{ \begin{array}{l}
x – y – 1 = 0\\
y + \frac{4}{3} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = – \frac{1}{3}\\
y = – \frac{4}{3}
\end{array} \right.\\
*)\\
F = \frac{3}{{2x – {x^2} – 4}}\\
2x – {x^2} – 4 = – \left( {{x^2} – 2x + 1} \right) – 3 = – {\left( {x – 1} \right)^2} – 3 \le – 3,\,\,\,\forall x\\
\Rightarrow F = \frac{3}{{ – {{\left( {x – 1} \right)}^2} – 3}} \ge \frac{3}{{ – 3}} = – 1\\
\Rightarrow {F_{\min }} = – 1 \Leftrightarrow x = 1\\
*)\\
G = \frac{2}{{6x – 5 – 9{x^2}}}\\
6x – 5 – 9{x^2} = – \left( {9{x^2} – 6x + 1} \right) – 4 = – 4 – {\left( {3x – 1} \right)^2} \le – 4\\
\Rightarrow G \ge \frac{2}{{ – 4}} = – \frac{1}{2}\\
\Rightarrow {G_{\min }} = – \frac{1}{2} \Leftrightarrow x = \frac{1}{3}
\end{array}\)