Đáp án:
\(\begin{array}{l}
a)\dfrac{{\sqrt x }}{{\sqrt x + 1}}\\
b)\dfrac{{3 - \sqrt 3 }}{6}\\
c)0 \le x < 4;x \ne 1\\
d)Max = 2
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ge 0;x \ne 1\\
A = \dfrac{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}} - \dfrac{{x + 1}}{{\sqrt x + 1}}\\
= \dfrac{{x + \sqrt x + 1}}{{\sqrt x + 1}} - \dfrac{{x + 1}}{{\sqrt x + 1}}\\
= \dfrac{{x + \sqrt x + 1 - x - 1}}{{\sqrt x + 1}}\\
= \dfrac{{\sqrt x }}{{\sqrt x + 1}}\\
b)Thay:x = 7 - 4\sqrt 3 \\
= 4 - 2.2.\sqrt 3 + 3\\
= {\left( {2 - \sqrt 3 } \right)^2}\\
\to A = \dfrac{{\sqrt {{{\left( {2 - \sqrt 3 } \right)}^2}} }}{{\sqrt {{{\left( {2 - \sqrt 3 } \right)}^2}} + 1}}\\
= \dfrac{{2 - \sqrt 3 }}{{2 - \sqrt 3 + 1}} = \dfrac{{2 - \sqrt 3 }}{{3 - \sqrt 3 }} = \dfrac{{3 - \sqrt 3 }}{6}\\
c)A < \dfrac{2}{3}\\
\to \dfrac{{\sqrt x }}{{\sqrt x + 1}} < \dfrac{2}{3}\\
\to \dfrac{{3\sqrt x - 2\sqrt x - 2}}{{3\left( {\sqrt x + 1} \right)}} < 0\\
\to \dfrac{{\sqrt x - 2}}{{3\left( {\sqrt x + 1} \right)}} < 0\\
\to \sqrt x - 2 < 0\left( {do:\sqrt x + 1 > 0\forall x \ge 0} \right)\\
\to 0 \le x < 4;x \ne 1\\
d)B = A - \sqrt x - 3\\
= \dfrac{{\sqrt x }}{{\sqrt x + 1}} - \sqrt x - 3\\
= \dfrac{{\sqrt x - x - \sqrt x - 3\sqrt x - 3}}{{\sqrt x + 1}}\\
= \dfrac{{ - x - 3\sqrt x - 3}}{{\sqrt x + 1}}\\
= \dfrac{{ - \left( {x + 2\sqrt x + 1} \right) - \sqrt x - 1}}{{\sqrt x + 1}}\\
= \dfrac{{ - {{\left( {\sqrt x + 1} \right)}^2} - \left( {\sqrt x + 1} \right)}}{{\sqrt x + 1}}\\
= - \left( {\sqrt x + 1} \right) - 1\\
= - \sqrt x - 1 - 1\\
= - \sqrt x + 2\\
Do:\sqrt x \ge 0\forall x \ge 0\\
\to - \sqrt x \le 0\\
\to - \sqrt x + 2 \le 2\\
\to Max = 2\\
\Leftrightarrow x = 0
\end{array}\)
