Toán Lớp 8: Tìm GTNN của
B=25x^2+3y^2-10xy+4y+1
C=(x+1)(x-2)(x-3)(x-6)
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Toán Lớp 8: Tìm GTNN của
B=25x^2+3y^2-10xy+4y+1
C=(x+1)(x-2)(x-3)(x-6)
Mình cần gấp
Bạn nào giúp mik mik sẽ vote 5 nhé
Comments ( 1 )
{B_{\min }} = – 1\\
{C_{\min }} = – 36
\end{array}\]
*)\\
B = 25{x^2} + 3{y^2} – 10xy + 4y + 1\\
= \left( {25{x^2} – 10xy + {y^2}} \right) + \left( {2{y^2} + 4y + 2} \right) – 1\\
= \left[ {{{\left( {5x} \right)}^2} – 2.5x.y + {y^2}} \right] + 2.\left( {{y^2} + 2y + 1} \right) – 1\\
= {\left( {5x – y} \right)^2} + 2.\left( {{y^2} + 2.y.1 + {1^2}} \right) – 1\\
= {\left( {5x – y} \right)^2} + 2.{\left( {y + 1} \right)^2} – 1\\
{\left( {5x – y} \right)^2} \ge 0,\,\,\,\forall x,y\\
{\left( {y + 1} \right)^2} \ge 0,\,\,\,\forall y\\
\Rightarrow B = {\left( {5x – y} \right)^2} + 2.{\left( {y + 1} \right)^2} – 1 \ge – 1,\,\,\,\forall x,y\\
\Rightarrow {B_{\min }} = – 1 \Leftrightarrow \left\{ \begin{array}{l}
{\left( {5x – y} \right)^2} = 0\\
{\left( {y + 1} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
5x – y = 0\\
y + 1 = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = \dfrac{{ – 1}}{5}\\
y = – 1
\end{array} \right.\\
\Rightarrow {B_{\min }} = – 1\\
*)\\
C = \left( {x + 1} \right)\left( {x – 2} \right)\left( {x – 3} \right)\left( {x – 6} \right)\\
= \left[ {\left( {x + 1} \right)\left( {x – 6} \right)} \right].\left[ {\left( {x – 2} \right)\left( {x – 3} \right)} \right]\\
= \left( {{x^2} – 6x + x – 6} \right).\left( {{x^2} – 3x – 2x + 6} \right)\\
= \left( {{x^2} – 5x – 6} \right)\left( {{x^2} – 5x + 6} \right)\\
= \left[ {\left( {{x^2} – 5x} \right) – 6} \right].\left[ {\left( {{x^2} – 5x} \right) + 6} \right]\\
= {\left( {{x^2} – 5x} \right)^2} – {6^2}\\
= {\left( {{x^2} – 5x} \right)^2} – 36\\
{\left( {{x^2} – 5x} \right)^2} \ge 0,\,\,\,\forall x\\
\Rightarrow C = {\left( {{x^2} – 5x} \right)^2} – 36 \ge – 36,\,\,\,\forall x\\
\Rightarrow {C_{\min }} = – 36 \Leftrightarrow {\left( {{x^2} – 5x} \right)^2} = 0 \Leftrightarrow {x^2} – 5x = 0\\
\Leftrightarrow x\left( {x – 5} \right) = 0 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x – 5 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 5
\end{array} \right.\\
\Rightarrow {C_{\min }} = – 36
\end{array}\)