Đáp án:
B9:
b) x>16
Giải thích các bước giải:
\(\begin{array}{l}
B8:\\
a)DK:x \ge 0;x \ne 1\\
B = \left[ {\dfrac{{2x + 1 - \sqrt x \left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}} \right].\left[ {\dfrac{{\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}}{{\sqrt x + 1}} - \sqrt x } \right]\\
= \dfrac{{2x + 1 - x + \sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}.\left( {x - \sqrt x + 1 - \sqrt x } \right)\\
= \dfrac{{x + \sqrt x + 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}.{\left( {\sqrt x - 1} \right)^2}\\
= \sqrt x - 1\\
b)B = 3\\
\to \sqrt x - 1 = 3\\
\to \sqrt x = 4\\
\to x = 16\\
B9)\\
a)DK:x > 0;x \ne 9\\
C = \left[ {\dfrac{{\sqrt x \left( {\sqrt x - 3} \right) - x - 9}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}} \right]:\left[ {\dfrac{{3\sqrt x + 1 - \sqrt x + 3}}{{\sqrt x \left( {\sqrt x - 3} \right)}}} \right]\\
= \dfrac{{x - 3\sqrt x - x - 9}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}.\dfrac{{\sqrt x \left( {\sqrt x - 3} \right)}}{{2\sqrt x + 4}}\\
= \dfrac{{ - 3\left( {\sqrt x + 3} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}.\dfrac{{\sqrt x \left( {\sqrt x - 3} \right)}}{{2\sqrt x + 4}}\\
= \dfrac{{ - 3\sqrt x }}{{2\sqrt x + 4}}\\
b)C < - 1\\
\to \dfrac{{ - 3\sqrt x }}{{2\sqrt x + 4}} < - 1\\
\to \dfrac{{3\sqrt x }}{{2\sqrt x + 4}} > 1\\
\to \dfrac{{3\sqrt x - 2\sqrt x - 4}}{{2\sqrt x + 4}} > 0\\
\to \dfrac{{\sqrt x - 4}}{{2\sqrt x + 4}} > 0\\
\to \sqrt x - 4 > 0\left( {do:2\sqrt x + 4 > 0\forall x > 0} \right)\\
\to x > 16
\end{array}\)
