Đáp án:
\(\begin{array}{l}
B1:\\
a) - 11\sqrt 5 \\
c)4\\
b)5\\
d)2\\
B2:\\
x = \dfrac{{51}}{{25}}\\
B3:\\
a)2\sqrt 3 + \sqrt 6 + 2 - 2\sqrt 2 \\
b)\dfrac{{ - 2\sqrt x }}{{\sqrt x + 1}}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
B1:\\
a)2\sqrt 5 - 2.3\sqrt 5 - 3.4\sqrt 5 + 5\sqrt 5 \\
= \left( {2 - 6 - 12 + 5} \right)\sqrt 5 \\
= - 11\sqrt 5 \\
c)\sqrt {9 + 2.3.\sqrt 5 + 5} - \left| {\sqrt 5 - 1} \right|\\
= \sqrt {{{\left( {3 + \sqrt 5 } \right)}^2}} - \sqrt 5 + 1\\
= 3 + \sqrt 5 - \sqrt 5 + 1\\
= 4\\
b)3\sqrt 2 .\left( {2 - \sqrt 2 } \right) + 2 - 6\sqrt 2 + 9\\
= 6\sqrt 2 - 3.2 + 11 - 6\sqrt 2 \\
= 5\\
d)\dfrac{{\sqrt 5 \left( {\sqrt 5 - 1} \right)}}{{\sqrt 5 - 1}} - \dfrac{{\sqrt 5 - 2}}{{5 - 4}}\\
= \sqrt 5 - \sqrt 5 + 2\\
= 2\\
B2:\\
DK:x \ge 2\\
\sqrt {16\left( {x - 2} \right)} - \sqrt {4\left( {x - 2} \right)} + \sqrt {9\left( {x - 2} \right)} = 1\\
\to 4\sqrt {x - 2} - 2\sqrt {x - 2} + 3\sqrt {x - 2} = 1\\
\to 5\sqrt {x - 2} = 1\\
\to \sqrt {x - 2} = \dfrac{1}{5}\\
\to x - 2 = \dfrac{1}{{25}}\\
\to x = \dfrac{{51}}{{25}}\\
B3:\\
a)\dfrac{{\left( {3 - \sqrt 2 } \right)\left( {\sqrt 3 - 2} \right)}}{{3 - 4}} + \dfrac{{6\left( {\sqrt 3 + 3} \right)}}{{3 - 9}} + \dfrac{{\sqrt 3 \left( {6\sqrt 3 - 1} \right)}}{{\sqrt 3 }}\\
= - \left( {3\sqrt 3 - 6 - \sqrt 6 + 2\sqrt 2 } \right) - \sqrt 3 - 3 + 6\sqrt 3 - 1\\
= - 3\sqrt 3 + 6 + \sqrt 6 - 2\sqrt 2 - \sqrt 3 - 3 + 6\sqrt 3 - 1\\
= 2\sqrt 3 + \sqrt 6 + 2 - 2\sqrt 2 \\
b)\dfrac{{\sqrt x \left( {\sqrt x - 2} \right) - x - 4}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}:\dfrac{{2\sqrt x - 1 - \sqrt x + 2}}{{\sqrt x \left( {\sqrt x - 2} \right)}}\\
= \dfrac{{x - 2\sqrt x - x - 4}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}.\dfrac{{\sqrt x \left( {\sqrt x - 2} \right)}}{{\sqrt x + 1}}\\
= \dfrac{{ - 2\left( {\sqrt x + 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}.\dfrac{{\sqrt x \left( {\sqrt x - 2} \right)}}{{\sqrt x + 1}}\\
= \dfrac{{ - 2\sqrt x }}{{\sqrt x + 1}}
\end{array}\)
