Đáp án:
B2:
f) 5
Giải thích các bước giải:
\(\begin{array}{l}
B1:\\
a)DK:2x - 1 \ge 0 \to x \ge \dfrac{1}{2}\\
b)DK:x \ge 0;x \ne 49\\
c)DK:2 - 6x > 0 \to \dfrac{1}{3} > x\\
d)DK:{x^2} - 4 \ge 0 \to \left[ \begin{array}{l}
x \ge 2\\
x \le - 2
\end{array} \right.
\end{array}\)
\(\begin{array}{l}
B2:\\
a)2\sqrt 3 + \left| {2 - \sqrt 3 } \right| = 2\sqrt 3 + 2 - \sqrt 3 = 2 + \sqrt 3 \\
c)2.\dfrac{4}{{\sqrt 3 }} - 3.\dfrac{1}{{3\sqrt 3 }} - 6.\dfrac{2}{{5\sqrt 3 }}\\
= \dfrac{8}{{\sqrt 3 }} - \dfrac{1}{{\sqrt 3 }} - \dfrac{{12}}{{5\sqrt 3 }} = \dfrac{{40 - 5 - 12}}{{5\sqrt 3 }} = \dfrac{{23}}{{5\sqrt 3 }}\\
e)\sqrt {2 + 2.\sqrt 2 .1 + 1} - \sqrt {4 - 2.2.\sqrt 2 + 2} \\
= \sqrt {{{\left( {\sqrt 2 + 1} \right)}^2}} - \sqrt {{{\left( {2 - \sqrt 2 } \right)}^2}} \\
= \sqrt 2 + 1 - 2 + \sqrt 2 = 2\sqrt 2 + 1\\
b)3\sqrt 2 - 4.3\sqrt 2 + 2.4\sqrt 2 - 5\sqrt 2 = - 6\sqrt 2 \\
d)\sqrt {\dfrac{{{{\left( {2 - \sqrt 3 } \right)}^2}}}{{4 - 3}}} + \sqrt {\dfrac{{{{\left( {2 + \sqrt 3 } \right)}^2}}}{{4 - 3}}} \\
= \dfrac{{2 - \sqrt 3 }}{1} + \dfrac{{2 + \sqrt 3 }}{1} = 4
\end{array}\)
\(\begin{array}{l}
f)\sqrt {5\sqrt 3 + 5\sqrt {48 - 10\sqrt {4 + 2.2.\sqrt 3 + 3} } } \\
= \sqrt {5\sqrt 3 + 5\sqrt {48 - 10\sqrt {{{\left( {2 + \sqrt 3 } \right)}^2}} } } \\
= \sqrt {5\sqrt 3 + 5\sqrt {48 - 10\left( {2 + \sqrt 3 } \right)} } \\
= \sqrt {5\sqrt 3 + 5\sqrt {28 - 10\sqrt 3 } } \\
= \sqrt {5\sqrt 3 + 5\sqrt {25 - 2.5\sqrt 3 + 3} } \\
= \sqrt {5\sqrt 3 + 5\sqrt {{{\left( {5 - \sqrt 3 } \right)}^2}} } \\
= \sqrt {5\sqrt 3 + 5\left( {5 - \sqrt 3 } \right)} \\
= \sqrt {25} = 5
\end{array}\)
