$+) \quad T = (3x - 2)^2 - (3 + 2x)^2$
$\to T = (3x - 2 - 3 + 2x)(3x - 2 + 3 + 2x)$
$\to T = (5x-5)(5x + 1)$
$\to T = 25x^2 + 20x - 5$
$\to T = (5x +2)^2 - 9$
Ta có:
$(5x + 2)^2 \geq o,\,\forall x$
$\to (5x +2)^2 - 9 \geq -9$
hay $T \geq - 9$
Dấu = xảy ra $\Leftrightarrow 5x + 2 = 0 \Leftrightarrow x = - \dfrac{2}{5}$
Vậy $\min T = - 9 \Leftrightarrow x = -\dfrac{2}{5}$
$+) \quad X = (5x - 6)^2 - 2(3x + 4)^2$
$\to X = 25x^2 - 60x + 36 - 2(9x^2 + 24x + 16)$
$\to X = 7x^2 -108x + 4$
$\to X = 7\left(x^2 - 2.\dfrac{54}{7}x + \dfrac{2916}{49}\right) - \dfrac{2888}{7}$
$\to X = 7\left(x - \dfrac{54}{7}\right)^2 - \dfrac{2888}{7}$
Ta có:
$\left(x - \dfrac{54}{7}\right)^2 \geq 0,\,\,\forall x$
$\to 7\left(x - \dfrac{54}{7}\right)^2 - \dfrac{2888}{7} \geq - \dfrac{2888}{7}$
$\to X \geq - \dfrac{2888}{7}$
Dấu = xảy ra $\Leftrightarrow x - \dfrac{54}{7} = 0 \Leftrightarrow x = \dfrac{54}{7}$
Vậy $\min X = -\dfrac{2888}{7} \Leftrightarrow x = \dfrac{54}{7}$
$+) \quad P = 3x^2 - 6x + 4$
$\to P = 3(x^2 - 2x + 1) + 1$
$\to P = 3(x-1)^2 + 1$
Ta có:
$(x-1)^2 \geq 0,\,\,\forall x$
$\to 3(x-1)^2 + 1 \geq 1$
$\to P \geq 1$
Dấu = xảy ra $\Leftrightarrow x - 1 = 0 \Leftrightarrow x = 1$
Vậy $\min P = 1 \Leftrightarrow x = 1$
