Đáp án:
B11:
a. \(\dfrac{{ - 3}}{{\sqrt x + 3}}\)
Giải thích các bước giải:
\(\begin{array}{l}
B11:\\
a.P = \left[ {\dfrac{{2\sqrt x \left( {\sqrt x - 3} \right) + \sqrt x \left( {\sqrt x + 3} \right) - 3x - 3}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}} \right]:\left( {\dfrac{{2\sqrt x - 2 - \sqrt x + 3}}{{\sqrt x - 3}}} \right)\\
= \left[ {\dfrac{{2x - 6\sqrt x + x + 3\sqrt x - 3x - 3}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}} \right].\dfrac{{\sqrt x - 3}}{{\sqrt x + 1}}\\
= \dfrac{{ - 3\sqrt x - 3}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}.\dfrac{{\sqrt x - 3}}{{\sqrt x + 1}} = \dfrac{{ - 3}}{{\sqrt x + 3}}\\
b.Thay:x = 4 - 2\sqrt 3 = {\left( {\sqrt 3 - 1} \right)^2}\\
\to P = \dfrac{{ - 3}}{{\sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} + 3}} = \dfrac{{ - 3}}{{\sqrt 3 - 1 + 3}}\\
= \dfrac{{ - 3}}{{2 + 2\sqrt 3 }}\\
c.P < - \dfrac{1}{2}\\
\to \dfrac{{ - 3}}{{\sqrt x + 3}} < - \dfrac{1}{2}\\
\to \dfrac{3}{{\sqrt x + 3}} > \dfrac{1}{2}\\
\to \dfrac{{6 - 2\sqrt x - 6}}{{2\left( {\sqrt x + 3} \right)}} > 0\\
\to \dfrac{{ - 2\sqrt x }}{{2\left( {\sqrt x + 3} \right)}} > 0\\
\to \sqrt x < 0\left( {vô lý} \right)\\
\left( {do:\sqrt x + 3 > 0\forall x \ge 0} \right)\\
\to x \in \emptyset \\
d.P \in Z\\
\Leftrightarrow \dfrac{3}{{\sqrt x + 3}} \in U\left( 3 \right)\\
Mà:\sqrt x + 3 \ge 3\forall x \ge 0\\
\to \sqrt x + 3 = 3\\
\Leftrightarrow x = 0\\
B12:\\
a.P = \dfrac{{x + 2 + \left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right) - \left( {x + \sqrt x + 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\\
= \dfrac{{x + 2 + x - 1 - x - \sqrt x - 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\\
= \dfrac{{x - \sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\\
= \dfrac{{\sqrt x }}{{x + \sqrt x + 1}}\\
b.Thay:x = 28 - 6\sqrt 3 = {\left( {3\sqrt 3 } \right)^2} - 2.3\sqrt 3 .1 + 1\\
= {\left( {3\sqrt 3 - 1} \right)^2}\\
\to P = \dfrac{{\sqrt {{{\left( {3\sqrt 3 - 1} \right)}^2}} }}{{28 - 6\sqrt 3 + \sqrt {{{\left( {3\sqrt 3 - 1} \right)}^2}} + 1}} = \dfrac{{3\sqrt 3 - 1}}{{28 - 6\sqrt 3 + 3\sqrt 3 - 1 + 1}}\\
= \dfrac{{3\sqrt 3 - 1}}{{28 - 3\sqrt 3 }}\\
c.P < \dfrac{1}{3}\\
\to \dfrac{{\sqrt x }}{{x + \sqrt x + 1}} < \dfrac{1}{3}\\
\to \dfrac{{3\sqrt x - x - \sqrt x - 1}}{{x + \sqrt x + 1}} < 0\\
\to - x + 2\sqrt x - 1 < 0\left( {do:x + \sqrt x + 1 > 0\forall x \ge 0} \right)\\
\to - {\left( {\sqrt x - 1} \right)^2} < 0\\
\to {\left( {\sqrt x - 1} \right)^2} > 0\\
\Leftrightarrow \sqrt x - 1 \ne 0\\
\Leftrightarrow x \ne 1;x \ge 0
\end{array}\]\[\begin{array}{l}
B11:\\
a.P = \left[ {\dfrac{{2\sqrt x \left( {\sqrt x - 3} \right) + \sqrt x \left( {\sqrt x + 3} \right) - 3x - 3}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}} \right]:\left( {\dfrac{{2\sqrt x - 2 - \sqrt x + 3}}{{\sqrt x - 3}}} \right)\\
= \left[ {\dfrac{{2x - 6\sqrt x + x + 3\sqrt x - 3x - 3}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}} \right].\dfrac{{\sqrt x - 3}}{{\sqrt x + 1}}\\
= \dfrac{{ - 3\sqrt x - 3}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}.\dfrac{{\sqrt x - 3}}{{\sqrt x + 1}} = \dfrac{{ - 3}}{{\sqrt x + 3}}\\
b.Thay:x = 4 - 2\sqrt 3 = {\left( {\sqrt 3 - 1} \right)^2}\\
\to P = \dfrac{{ - 3}}{{\sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} + 3}} = \dfrac{{ - 3}}{{\sqrt 3 - 1 + 3}}\\
= \dfrac{{ - 3}}{{2 + 2\sqrt 3 }}\\
c.P < - \dfrac{1}{2}\\
\to \dfrac{{ - 3}}{{\sqrt x + 3}} < - \dfrac{1}{2}\\
\to \dfrac{3}{{\sqrt x + 3}} > \dfrac{1}{2}\\
\to \dfrac{{6 - 2\sqrt x - 6}}{{2\left( {\sqrt x + 3} \right)}} > 0\\
\to \dfrac{{ - 2\sqrt x }}{{2\left( {\sqrt x + 3} \right)}} > 0\\
\to \sqrt x < 0\left( {voly} \right)\\
\left( {do:\sqrt x + 3 > 0\forall x \ge 0} \right)\\
\to x \in \emptyset \\
d.P \in Z\\
\Leftrightarrow \dfrac{3}{{\sqrt x + 3}} \in U\left( 3 \right)\\
Ma:\sqrt x + 3 \ge 3\forall x \ge 0\\
\to \sqrt x + 3 = 3\\
\Leftrightarrow x = 0\\
B12:\\
a.P = \dfrac{{x + 2 + \left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right) - \left( {x + \sqrt x + 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\\
= \dfrac{{x + 2 + x - 1 - x - \sqrt x - 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\\
= \dfrac{{x - \sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\\
= \dfrac{{\sqrt x }}{{x + \sqrt x + 1}}\\
b.Thay:x = 28 - 6\sqrt 3 = {\left( {3\sqrt 3 } \right)^2} - 2.3\sqrt 3 .1 + 1\\
= {\left( {3\sqrt 3 - 1} \right)^2}\\
\to P = \dfrac{{\sqrt {{{\left( {3\sqrt 3 - 1} \right)}^2}} }}{{28 - 6\sqrt 3 + \sqrt {{{\left( {3\sqrt 3 - 1} \right)}^2}} + 1}} = \dfrac{{3\sqrt 3 - 1}}{{28 - 6\sqrt 3 + 3\sqrt 3 - 1 + 1}}\\
= \dfrac{{3\sqrt 3 - 1}}{{28 - 3\sqrt 3 }}\\
c.P < \dfrac{1}{3}\\
\to \dfrac{{\sqrt x }}{{x + \sqrt x + 1}} < \dfrac{1}{3}\\
\to \dfrac{{3\sqrt x - x - \sqrt x - 1}}{{x + \sqrt x + 1}} < 0\\
\to - x + 2\sqrt x - 1 < 0\left( {do:x + \sqrt x + 1 > 0\forall x \ge 0} \right)\\
\to - {\left( {\sqrt x - 1} \right)^2} < 0\\
\to {\left( {\sqrt x - 1} \right)^2} > 0\\
\Leftrightarrow \sqrt x - 1 \ne 0\\
\Leftrightarrow x \ne 1;x \ge 0
\end{array}\)
