Đáp án:
\(\begin{array}{l}
B6:\\
a)\dfrac{{\sqrt a - 1}}{{\sqrt a }}\\
b)M < 1\\
B7:\\
1)x - \sqrt x + 3\\
2)Min = \dfrac{{11}}{4}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
B6:\\
a)DK:a > 0;a \ne 1\\
M = \dfrac{{1 + \sqrt a }}{{\sqrt a \left( {\sqrt a - 1} \right)}}:\dfrac{{\sqrt a + 1}}{{{{\left( {\sqrt a - 1} \right)}^2}}}\\
= \dfrac{{1 + \sqrt a }}{{\sqrt a \left( {\sqrt a - 1} \right)}}.\dfrac{{{{\left( {\sqrt a - 1} \right)}^2}}}{{\sqrt a + 1}}\\
= \dfrac{{\sqrt a - 1}}{{\sqrt a }}\\
b)M - 1 = \dfrac{{\sqrt a - 1}}{{\sqrt a }} - 1 = \dfrac{{\sqrt a - 1 - \sqrt a }}{{\sqrt a }}\\
= - \dfrac{1}{{\sqrt a }}\\
Do:\dfrac{1}{{\sqrt a }} > 0\forall a > 0\\
\to - \dfrac{1}{{\sqrt a }} < 0\\
\to M < 1\\
B7:\\
1)P = \dfrac{{\sqrt x \left( {\sqrt x - 2} \right) + \sqrt x + 2 + 1}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}.\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)\\
= x - \sqrt x + 3\\
2)P = x - 2\sqrt x .\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{{11}}{4}\\
= {\left( {\sqrt x - \dfrac{1}{2}} \right)^2} + \dfrac{{11}}{4}\\
Do:{\left( {\sqrt x - \dfrac{1}{2}} \right)^2} \ge 0\forall x \ge 0\\
\to {\left( {\sqrt x - \dfrac{1}{2}} \right)^2} + \dfrac{{11}}{4} \ge \dfrac{{11}}{4}\\
\to Min = \dfrac{{11}}{4}\\
\Leftrightarrow x = \dfrac{1}{4}
\end{array}\)
