Đáp án:
\(\begin{array}{l}
B1:\\
1)A = 0\\
2)B = 4\sqrt 3 - 4\\
B2:\\
1)C = \dfrac{{\sqrt x + 2}}{{\sqrt x }}\\
2)x = \dfrac{{16}}{9}\\
B3:\\
1)x = 4\\
2)\left[ \begin{array}{l}
x = 3\\
x = 6
\end{array} \right.
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
B1:\\
1)A = \left( {\dfrac{1}{2}.2\sqrt 7 - 2\sqrt 3 - \sqrt 7 } \right).\sqrt 7 + 2\sqrt {21} \\
= \left( { - 2\sqrt 3 } \right)\sqrt 7 + 2\sqrt {21} \\
= - 2\sqrt {21} + 2\sqrt {21} = 0\\
2)B = 3 + 2\sqrt 3 + 1 + 2\left( {\sqrt 3 - 2} \right) - 4.\left( {\dfrac{{\sqrt 3 + 1 - \sqrt 3 + 1}}{{3 - 1}}} \right)\\
= 4 + 2\sqrt 3 + 2\sqrt 3 - 4 - 4.\dfrac{2}{2}\\
= 4\sqrt 3 - 4\\
B2:\\
1)C = \dfrac{{1 - \sqrt x }}{{\sqrt x \left( {\sqrt x + 2} \right)}}:\dfrac{{1 - \sqrt x }}{{{{\left( {\sqrt x + 2} \right)}^2}}}\\
= \dfrac{{1 - \sqrt x }}{{\sqrt x \left( {\sqrt x + 2} \right)}}.\dfrac{{{{\left( {\sqrt x + 2} \right)}^2}}}{{1 - \sqrt x }}\\
= \dfrac{{\sqrt x + 2}}{{\sqrt x }}\\
2)C = \dfrac{5}{2}\\
\to \dfrac{{\sqrt x + 2}}{{\sqrt x }} = \dfrac{5}{2}\\
\to 2\sqrt x + 4 = 5\sqrt x \\
\to 3\sqrt x = 4\\
\to \sqrt x = \dfrac{4}{3}\\
\to x = \dfrac{{16}}{9}\\
B3:\\
1)DK:x \ge 0\\
x - 4\sqrt x + 4 = 0\\
\to {\left( {\sqrt x - 2} \right)^2} = 0\\
\to \sqrt x - 2 = 0\\
\to x = 4\\
2)DK:x \ge 3\\
\sqrt {\left( {x - 3} \right)\left( {x + 3} \right)} - 3\sqrt {x - 3} = 0\\
\to \sqrt {x - 3} \left( {\sqrt {x + 3} - 3} \right) = 0\\
\to \left[ \begin{array}{l}
x - 3 = 0\\
\sqrt {x + 3} = 3
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 3\\
x + 3 = 9
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 3\\
x = 6
\end{array} \right.
\end{array}\)
