Đáp án:
i. 1
Giải thích các bước giải:
\(\begin{array}{l}
f.\dfrac{{\sqrt 5 + \sqrt 3 }}{{5 - 3}} + \dfrac{{\sqrt {15} \left( {\sqrt 5 - \sqrt 3 } \right)}}{{2\sqrt {15} }} - 2\sqrt 5 \\
= \dfrac{{\sqrt 5 + \sqrt 3 }}{2} + \dfrac{{\sqrt 5 - \sqrt 3 }}{2} - 2\sqrt 5 \\
= \dfrac{{2\sqrt 5 }}{2} - 2\sqrt 5 \\
= \sqrt 5 - 2\sqrt 5 = - \sqrt 5 \\
g.\dfrac{{\sqrt {4 - 2.2.\sqrt 3 + 3} }}{{\sqrt {2 - \sqrt 3 } }} + \sqrt {2 + \sqrt 3 } \\
= \dfrac{{\sqrt {{{\left( {2 - \sqrt 3 } \right)}^2}} + \sqrt {\left( {2 + \sqrt 3 } \right)\left( {2 - \sqrt 3 } \right)} }}{{\sqrt {2 - \sqrt 3 } }}\\
= \dfrac{{2 - \sqrt 3 + \sqrt {4 - 3} }}{{\sqrt {2 - \sqrt 3 } }}\\
= \dfrac{{2 - \sqrt 3 + 1}}{{\sqrt {2 - \sqrt 3 } }}\\
= \dfrac{{3 - \sqrt 3 }}{{\sqrt {2 - \sqrt 3 } }}\\
= \dfrac{{3\sqrt 2 - \sqrt 6 }}{{\sqrt {4 - 2\sqrt 3 } }}\\
= \dfrac{{3\sqrt 2 - \sqrt 6 }}{{\sqrt {3 - 2\sqrt 3 .1 + 1} }}\\
= \dfrac{{3\sqrt 2 - \sqrt 6 }}{{\sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} }}\\
= \dfrac{{\sqrt 6 \left( {\sqrt 3 - 1} \right)}}{{\sqrt 3 - 1}} = \sqrt 6 \\
i.\dfrac{{2\left( {2\sqrt 3 + 2} \right) - 2\left( {2\sqrt 3 - 2} \right)}}{{\left( {2\sqrt 3 + 2} \right)\left( {2\sqrt 3 - 2} \right)}}\\
= \dfrac{{4\sqrt 3 + 4 - 4\sqrt 3 + 4}}{{12 - 4}}\\
= \dfrac{8}{8} = 1
\end{array}\)
( câu h bạn xem lại đề nhé )
