Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
A = {x^2} - 8x + 5 = \left( {{x^2} - 8x + 16} \right) - 11\\
= \left( {{x^2} - 2.x.4 + {4^2}} \right) - 11 = {\left( {x - 4} \right)^2} - 11\\
{\left( {x - 4} \right)^2} \ge 0,\,\,\,\forall x \Rightarrow {\left( {x - 4} \right)^2} - 11 \ge - 11,\,\,\,\forall x\\
\Rightarrow A \ge - 11,\,\,\,\forall x\\
\Rightarrow {A_{\min }} = - 11 \Leftrightarrow {\left( {x - 4} \right)^2} = 0 \Leftrightarrow x = 4\\
b,\\
B = 4{x^2} - 4x = \left( {4{x^2} - 4x + 1} \right) - 1\\
= \left[ {{{\left( {2x} \right)}^2} - 2.2x.1 + {1^2}} \right] - 1 = {\left( {2x - 1} \right)^2} - 1\\
{\left( {2x - 1} \right)^2} \ge 0,\,\,\,\forall x\\
\Rightarrow {\left( {2x - 1} \right)^2} - 1 \ge - 1,\,\,\,\forall x\\
\Rightarrow {B_{\min }} = - 1 \Leftrightarrow {\left( {2x - 1} \right)^2} = 0 \Leftrightarrow x = \dfrac{1}{2}\\
c,\\
C = {x^2} + x + 1 = \left( {{x^2} + x + \dfrac{1}{4}} \right) + \dfrac{3}{4}\\
= \left( {{x^2} + 2.x.\dfrac{1}{2} + {{\left( {\dfrac{1}{2}} \right)}^2}} \right) + \dfrac{3}{4} = {\left( {x + \dfrac{1}{2}} \right)^2} + \dfrac{3}{4}\\
{\left( {x + \dfrac{1}{2}} \right)^2} \ge 0,\,\,\,\forall x\\
\Rightarrow {\left( {x + \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} \ge \dfrac{3}{4},\,\,\,\forall x\\
\Rightarrow C \ge \dfrac{3}{4},\,\,\,\forall x\\
\Rightarrow {C_{\min }} = \dfrac{3}{4} \Leftrightarrow {\left( {x + \dfrac{1}{2}} \right)^2} = 0 \Leftrightarrow x = - \dfrac{1}{2}\\
d,\\
D = 2{x^2} + 6x - 4 = 2.\left( {{x^2} + 3x - 2} \right)\\
= 2.\left[ {\left( {{x^2} + 3x + \dfrac{9}{4}} \right) - \dfrac{{17}}{4}} \right]\\
= 2.\left[ {{{\left( {x + \dfrac{3}{2}} \right)}^2} - \dfrac{{17}}{4}} \right]\\
= 2.{\left( {x + \dfrac{3}{2}} \right)^2} - \dfrac{{17}}{2}\\
{\left( {x + \dfrac{3}{2}} \right)^2} \ge 0,\,\,\,\forall x\\
\Rightarrow 2.{\left( {x + \dfrac{3}{2}} \right)^2} - \dfrac{{17}}{2} \ge - \dfrac{{17}}{2},\,\,\,\forall x\\
\Rightarrow D \ge - \dfrac{{17}}{2},\,\,\,\forall x\\
\Rightarrow {D_{\min }} = - \dfrac{{17}}{2} \Leftrightarrow {\left( {x + \dfrac{3}{2}} \right)^2} = 0 \Leftrightarrow x = - \dfrac{3}{2}\\
e,\\
E = {x^2} + {y^2} - 6x + 5y + 1\\
= \left( {{x^2} - 6x + 9} \right) + \left( {{y^2} + 5y + \dfrac{{25}}{4}} \right) - \dfrac{{57}}{4}\\
= \left( {{x^2} - 2.x.3 + {3^3}} \right) + \left( {{y^2} + 2.y.\dfrac{5}{2} + {{\left( {\dfrac{5}{2}} \right)}^2}} \right) - \dfrac{{57}}{2}\\
= {\left( {x - 3} \right)^2} + {\left( {y + \dfrac{5}{2}} \right)^2} - \dfrac{{57}}{2}\\
{\left( {x - 3} \right)^2} \ge 0,\,\,\,\forall x\\
{\left( {y + \dfrac{5}{2}} \right)^2} \ge 0,\,\,\,\forall y\\
\Rightarrow {\left( {x - 3} \right)^2} + {\left( {y + \dfrac{5}{2}} \right)^2} - \dfrac{{57}}{2} \ge - \dfrac{{57}}{2},\,\,\,\forall x,y\\
\Rightarrow {E_{\min }} = - \dfrac{{57}}{2} \Leftrightarrow \left\{ \begin{array}{l}
{\left( {x - 3} \right)^2} = 0\\
{\left( {y + \dfrac{5}{2}} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 3\\
y = - \dfrac{5}{2}
\end{array} \right.\\
g,\\
G = 2{x^2} + {y^2} + 10x - 2xy + 26\\
= \left( {{x^2} - 2xy + {y^2}} \right) + \left( {{x^2} + 10x + 25} \right) + 1\\
= \left( {{x^2} - 2.x.y + {y^2}} \right) + \left( {{x^2} + 2.x.5 + {5^2}} \right) + 1\\
= {\left( {x - y} \right)^2} + {\left( {x + 5} \right)^2} + 1\\
{\left( {x - y} \right)^2} \ge 0,\,\,\,\forall x,y\\
{\left( {x + 5} \right)^2} \ge 0,\,\,\,\forall x\\
\Rightarrow {\left( {x - y} \right)^2} + {\left( {x + 5} \right)^2} + 1 \ge 1,\,\,\,\forall x,y\\
\Rightarrow {G_{\min }} = 1 \Leftrightarrow \left\{ \begin{array}{l}
{\left( {x - y} \right)^2} = 0\\
{\left( {x + 5} \right)^2} = 0
\end{array} \right. \Leftrightarrow x = y = - 5\\
h,\\
H = \left( {a - 1} \right)\left( {a - 2} \right)\left( {a - 3} \right)\left( {a - 4} \right)\\
= \left[ {\left( {a - 1} \right)\left( {a - 4} \right)} \right].\left[ {\left( {a - 2} \right).\left( {a - 3} \right)} \right]\\
= \left( {{a^2} - 5a + 4} \right).\left( {{a^2} - 5a + 6} \right)\\
= \left[ {\left( {{a^2} - 5a + 5} \right) - 1} \right].\left[ {\left( {{a^2} - 5a + 5} \right) + 1} \right]\\
= {\left( {{a^2} - 5a + 5} \right)^2} - 1\\
{\left( {{a^2} - 5a + 5} \right)^2} \ge 0,\,\,\,\forall a\\
\Rightarrow {\left( {{a^2} - 5a + 5} \right)^2} - 1 \ge - 1,\,\,\,\forall a\\
\Rightarrow {H_{\min }} = - 1 \Leftrightarrow {\left( {{a^2} - 5a + 5} \right)^2} = 0 \Leftrightarrow a = \dfrac{{5 \pm \sqrt 5 }}{2}
\end{array}\)
