Đáp án:
B4:
c. \(x > 1;x \ne 4\)
Giải thích các bước giải:
\(\begin{array}{l}
B3:DK:x \ge 0;x \ne 9\\
P = \dfrac{{x - 3\sqrt x + 2x + 6\sqrt x - 3x - 9}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{3\sqrt x - 9}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{3}{{\sqrt x + 3}}
\end{array}\)
Để P đạt GTNN
⇔ \({\sqrt x + 3}\) đạt GTLN
\(\begin{array}{l}
\Leftrightarrow \sqrt x + 3 = 9\\
\Leftrightarrow \sqrt x = 6\\
\to x = 36\\
\to MinP = \dfrac{3}{{\sqrt {36} + 3}} = \dfrac{1}{3}
\end{array}\)
\(\begin{array}{l}
B4:\\
a.DK:x > 0;x \ne 4\\
b.P = \left[ {\dfrac{{\sqrt x - 4 + 3\sqrt x }}{{\sqrt x \left( {\sqrt x - 2} \right)}}} \right]:\left[ {\dfrac{{x - 4 - x}}{{\sqrt x \left( {\sqrt x - 2} \right)}}} \right]\\
= \dfrac{{4\sqrt x - 4}}{{\sqrt x \left( {\sqrt x - 2} \right)}}.\dfrac{{\sqrt x \left( {\sqrt x - 2} \right)}}{{ - 4}}\\
= 1 - \sqrt x \\
Thay:x = 57 - 40\sqrt 2 = {\left( {4\sqrt 2 } \right)^2} - 2.4\sqrt 2 .5 + 25\\
= {\left( {4\sqrt 2 - 5} \right)^2}\\
\to P = 1 - \sqrt {{{\left( {4\sqrt 2 - 5} \right)}^2}} = 1 - 4\sqrt 2 + 5 = 6 - 4\sqrt 2 \\
c.P < 0\\
\to 1 - \sqrt x < 0\\
\to 1 < \sqrt x \\
\to x > 1;x \ne 4
\end{array}\)
