Tính GTLN, GTNN của các biểu thức ( nếu có)
F = (x – 3) ² + (x – 11) ² G = (x – 1)(x + 3)(x + 2)(x + 6) H = (x + 1)(x – 2)(x – 3)(x – 6)
I = 5 – 8x – x ²
pls giúp em vs ạ 🙁
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
F = {\left( {x – 3} \right)^2} + {\left( {x – 11} \right)^2}\\
= \left( {{x^2} – 6x + 9} \right) + \left( {{x^2} – 22x + 121} \right)\\
= 2{x^2} – 28x + 130\\
= 2.\left( {{x^2} – 14x + 49} \right) + 32\\
= 2.{\left( {x – 7} \right)^2} + 32 \ge 32,\,\,\,\forall x\\
\Rightarrow {F_{\min }} = 32 \Leftrightarrow {\left( {x – 7} \right)^2} = 0 \Leftrightarrow x = 7\\
G = \left( {x – 1} \right)\left( {x + 3} \right)\left( {x + 2} \right)\left( {x + 6} \right)\\
= \left[ {\left( {x – 1} \right)\left( {x + 6} \right)} \right].\left[ {\left( {x + 3} \right)\left( {x + 2} \right)} \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 \ge – 36,\,\,\,\forall x\\
\Rightarrow {G_{\min }} = 0 \Leftrightarrow {\left( {{x^2} + 5x} \right)^2} = 0 \Leftrightarrow {x^2} + 5x = 0 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = – 5
\end{array} \right.\\
H = \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} – 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 \ge – 36,\,\,\,\forall x\\
\Rightarrow {H_{\min }} = – 36 \Leftrightarrow {\left( {{x^2} – 5x} \right)^2} = 0 \Leftrightarrow {x^2} – 5x = x\left( {x – 5} \right) = 0 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 5
\end{array} \right.\\
I = 5 – 8x – {x^2}\\
= – \left( {{x^2} + 8x + 16} \right) + 21\\
= 21 – \left( {{x^2} + 2.x.4 + {4^2}} \right)\\
= 21 – {\left( {x + 4} \right)^2} \le 21,\,\,\,\forall x\\
\Rightarrow {I_{\max }} = 21 \Leftrightarrow {\left( {x + 4} \right)^2} = 0 \Leftrightarrow x = – 4
\end{array}\)