Đáp án:
C23:
b) \(\sqrt M = \sqrt 3 - 1\)
Giải thích các bước giải:
\(\begin{array}{l}
B22:\\
a)P = \dfrac{{\sqrt x \left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}}{{x - \sqrt x + 1}} - \dfrac{{\sqrt x \left( {2\sqrt x + 1} \right)}}{{\sqrt x }} + 1\\
= \sqrt x \left( {\sqrt x + 1} \right) - \left( {2\sqrt x + 1} \right) + 1\\
= x + \sqrt x - 2\sqrt x - 1 + 1\\
= x - \sqrt x \\
b)P = 2\\
\to x - \sqrt x = 2\\
\to \left[ \begin{array}{l}
\sqrt x = 1\\
\sqrt x = - 2\left( l \right)
\end{array} \right.\\
\to x = 1\\
c)P = x - \sqrt x = x - 2\sqrt x .\dfrac{1}{2} + \dfrac{1}{4} - \dfrac{1}{4}\\
= {\left( {\sqrt x - \dfrac{1}{2}} \right)^2} - \dfrac{1}{4}\\
Do:{\left( {\sqrt x - \dfrac{1}{2}} \right)^2} \ge 0\forall x > 0\\
\to {\left( {\sqrt x - \dfrac{1}{2}} \right)^2} - \dfrac{1}{4} \ge - \dfrac{1}{4}\\
\to Min = - \dfrac{1}{4}\\
\Leftrightarrow \sqrt x - \dfrac{1}{2} = 0\\
\Leftrightarrow x = \dfrac{1}{4}\\
C23:\\
a)M = \left[ {1 + \dfrac{{\sqrt x \left( {\sqrt x + 1} \right)}}{{\sqrt x + 1}}} \right].\left[ {1 - \dfrac{{\sqrt x \left( {\sqrt x - 1} \right)}}{{\sqrt x - 1}}} \right]\\
= \left( {1 + \sqrt x } \right)\left( {1 - \sqrt x } \right)\\
= 1 - x\\
b)Thay:x = - 3 + 2\sqrt 3 \\
\to M = 1 + 3 - 2\sqrt 3 = 3 - 2\sqrt 3 .1 + 1\\
= {\left( {\sqrt 3 - 1} \right)^2}\\
\to \sqrt M = \sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} = \sqrt 3 - 1
\end{array}\)
