Đáp án:
B6:
b. \(A = \dfrac{1}{2}\)
Giải thích các bước giải:
\(\begin{array}{l}
B5:\\
a.DK:x \ge 0;x \ne 9\\
B = \left[ {\dfrac{{2\sqrt x \left( {\sqrt x - 3} \right) + \sqrt x \left( {\sqrt x + 3} \right) - 3x - 3}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}} \right]:\left( {\dfrac{{2\sqrt x - 2 - \sqrt x + 3}}{{\sqrt x - 3}}} \right)\\
= \dfrac{{2x - 6\sqrt x + x + 3\sqrt x - 3x - 3}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}.\dfrac{{\sqrt x - 3}}{{\sqrt x + 1}}\\
= \dfrac{{ - 3\sqrt x - 3}}{{\sqrt x + 3}}.\dfrac{1}{{\sqrt x + 1}}\\
= - \dfrac{3}{{\sqrt x + 3}}\\
b.B < - \dfrac{1}{2}\\
\to - \dfrac{3}{{\sqrt x + 3}} < - \dfrac{1}{2}\\
\to \dfrac{3}{{\sqrt x + 3}} > \dfrac{1}{2}\\
\to \dfrac{{6 - \sqrt x - 3}}{{2\left( {\sqrt x + 3} \right)}} > 0\\
\to \dfrac{{3 - \sqrt x }}{{2\left( {\sqrt x + 3} \right)}} > 0\\
\to 3 - \sqrt x > 0\left( {Do:\sqrt x + 3 > 0\forall x \ge 0;x \ne 9} \right)\\
\to 9 > x\\
B6:\\
a.DK:x > 0;x \ne 9\\
A = \left[ {\dfrac{{x + 3 + \sqrt x - 3}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}} \right]:\dfrac{{\sqrt x }}{{\sqrt x - 3}}\\
= \dfrac{{x + \sqrt x }}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}.\dfrac{{\sqrt x - 3}}{{\sqrt x }}\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x + 3}}\\
b.Thay:x = \sqrt {27 + 10\sqrt 2 } - \sqrt {18 + 8\sqrt 2 } \\
= \sqrt {25 + 2.5.\sqrt 2 + 2} - \sqrt {16 + 2.4.\sqrt 2 + 2} \\
= \sqrt {{{\left( {5 + \sqrt 2 } \right)}^2}} - \sqrt {{{\left( {4 + \sqrt 2 } \right)}^2}} \\
= 5 + \sqrt 2 - 4 - \sqrt 2 = 1\\
\to A = \dfrac{{\sqrt 1 + 1}}{{\sqrt 1 + 3}} = \dfrac{2}{4} = \dfrac{1}{2}
\end{array}\)
