Đáp án:
a) \(Min = 1\)
Giải thích các bước giải:
\(\begin{array}{*{20}{l}}
{a)DK:x \ge 0}\\
{A = \dfrac{{2\left( {\sqrt x {\rm{\;}} + 1} \right) - 1}}{{\sqrt x {\rm{\;}} + 1}} = 2 - \dfrac{1}{{\sqrt x {\rm{\;}} + 1}}}\\
{Do:x \ge 0\forall x \ge 0}\\
{n{\rm{\;}} \to \sqrt x {\rm{\;}} \ge 0}\\
{n{\rm{\;}} \to \sqrt x {\rm{\;}} + 1 \ge 1}\\
{n{\rm{\;}} \to \dfrac{1}{{\sqrt x {\rm{\;}} + 1}} \le 1}\\
{n{\rm{\;}} \to {\rm{\;}} - \dfrac{1}{{\sqrt x {\rm{\;}} + 1}} \ge {\rm{\;}} - 1}\\
{n{\rm{\;}} \to 2 - \dfrac{1}{{\sqrt x {\rm{\;}} + 1}} \ge 1}\\
{n{\rm{\;}} \to Min = 1}\\
{n{\rm{\;}} \Leftrightarrow x = 0}\\
{b)DK:x \ge \dfrac{3}{2}}\\
{B = x - 1 - \sqrt {2x - 3} }\\
{ \to 2B = 2x - 2 - 2\sqrt {2x - 3} }\\
{ = 2x - 3 - 2.\sqrt {2x - 3} .1 + 1}\\
{ = {{\left( {\sqrt {2x - 3} {\rm{\;}} - 1} \right)}^2}}\\
{Do:\sqrt {2x - 3} {\rm{\;}} \ge 0\forall x \ge \dfrac{3}{2}}\\
{ \to \sqrt {2x - 3} {\rm{\;}} - 1 \ge {\rm{\;}} - 1}\\
{ \to {{\left( {\sqrt {2x - 3} {\rm{\;}} - 1} \right)}^2} \ge 1}\\
{ \to Min2B = 1 \to MinB = \dfrac{1}{2}}\\
{ \Leftrightarrow x = \dfrac{3}{2}}
\end{array}\)
