Đáp án:
\(\begin{array}{l}
a)\quad\min A = \dfrac{23}{16}\Leftrightarrow x = -\dfrac{5}{8}\\
b)\quad\min B = 2 \Leftrightarrow x = -4
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
\quad\ A = 4x^2 + 5x + 3\\
\Leftrightarrow A = 4\left(x^2 + \dfrac54x + \dfrac34\right)\\
\Leftrightarrow A = 4\left(x^2 + 2\cdot \dfrac58x + \dfrac{25}{64} + \dfrac{23}{64}\right)\\
\Leftrightarrow A = 4\left(x + \dfrac58\right)^2 + \dfrac{23}{16}\\
\text{Ta có:}\\
\quad \left(x + \dfrac58\right)^2 \geqslant 0\quad \forall x\\
\Leftrightarrow 4\left(x + \dfrac58\right)^2 + \dfrac{23}{16} \geqslant \dfrac{23}{16}\\
\Leftrightarrow A \geqslant \dfrac{23}{16}\\
\text{Dấu = xảy ra}\Leftrightarrow x + \dfrac{5}{8} = 0 \Leftrightarrow x = -\dfrac{5}{8}\\
\text{Vậy}\ \min A = \dfrac{23}{16}\Leftrightarrow x = -\dfrac{5}{8}\\
b)\\
\quad \ B = (x+3)^2 + (x+5)^2\\
\Leftrightarrow B = 2x^2 + 16x + 34\\
\Leftrightarrow B = 2(x^2 + 8x + 16 + 1)\\
\Leftrightarrow B = 2(x + 4)^2 + 2\\
\text{Ta có:}\\
\quad (x+4)^2 \geqslant 0\quad \forall x\\
\Leftrightarrow 2(x+4)^2 + 2 \geqslant 2\\
\Leftrightarrow B \geqslant 2\\
\text{Dấu = xảy ra}\Leftrightarrow x + 4 = 0 \Leftrightarrow x = -4\\
\text{Vậy}\ \min B = 2 \Leftrightarrow x = -4
\end{array}\)
