Đáp án:

\(\eqalign{ & a)\,\,GTNN\,\,A\,\,bang\,\,0 \Leftrightarrow x = {{9 \pm \sqrt {37} } \over 2} \cr & b)\,\,GTNN\,\,B\,\,bang\,\,1 \Leftrightarrow x = {1 \over 4} \cr & c)\,\,GTNN\,\,C\,\,bang\,\,{{\sqrt {10} } \over 2} \Leftrightarrow x = {1 \over 2} \cr} \)

Giải thích các bước giải:

$$\eqalign{ & a)\,\,A = \sqrt {{x^2} - 9x + 11} \cr & DKXD:\,\,{x^2} - 9x + 11 \ge 0 \Leftrightarrow \left[ \matrix{ x \ge {{9 + \sqrt {37} } \over 2} \hfill \cr x \le {{9 - \sqrt {37} } \over 2} \hfill \cr} \right. \cr & {x^2} - 9x + 11 = {x^2} - 2.x.{9 \over 2} + {{81} \over 4} - {{37} \over 4} \cr & = {\left( {x - {9 \over 2}} \right)^2} - {{37} \over 4} \cr & Ta\,\,co:\,\, \cr & + )\,\,x \ge {{9 + \sqrt {37} } \over 2} \Leftrightarrow x - {9 \over 2} \ge {{\sqrt {37} } \over 2} \cr & + )\,\,x \le {{9 - \sqrt {37} } \over 2} \Leftrightarrow x - {9 \over 2} \le {{ - \sqrt {37} } \over 2} \cr & \Leftrightarrow {\left( {x - {9 \over 2}} \right)^2} \ge {{37} \over 4} \Leftrightarrow {\left( {x - {9 \over 2}} \right)^2} - {{37} \over 4} \ge 0 \cr & \Rightarrow A \ge 0 \cr & Vay\,\,GTNN\,\,A\,\,bang\,\,0 \Leftrightarrow x = {{9 \pm \sqrt {37} } \over 2} \cr & b)\,\,B = \sqrt {4{x^2} - 4x + 2} \cr & B = \sqrt {{{\left( {4x - 1} \right)}^2} + 1} \ge 1 \cr & \Rightarrow GTNN\,\,B\,\,bang\,\,1 \Leftrightarrow x = {1 \over 4} \cr & c)\,\,C = \sqrt {2{x^2} - 2x + 3} \cr & C = \sqrt 2 \sqrt {{x^2} - x + {3 \over 2}} \cr & = \sqrt 2 \sqrt {{x^2} - 2.x.{1 \over 2} + {1 \over 4} + {5 \over 4}} \cr & = \sqrt 2 \sqrt {{{\left( {x - {1 \over 2}} \right)}^2} + {5 \over 4}} \cr & {\left( {x - {1 \over 2}} \right)^2} \ge 0 \Rightarrow {\left( {x - {1 \over 2}} \right)^2} + {5 \over 4} \ge {5 \over 4} \cr & \Rightarrow \sqrt {{{\left( {x - {1 \over 2}} \right)}^2} + {5 \over 4}} \ge {{\sqrt 5 } \over 2} \cr & \Rightarrow C \ge {{\sqrt {10} } \over 2} \cr & \Rightarrow GTNN\,\,C\,\,bang\,\,{{\sqrt {10} } \over 2} \Leftrightarrow x = {1 \over 2} \cr} $$