Đáp án:
6A.b. 1
Giải thích các bước giải:
\(\begin{array}{l}
3A.a.\sqrt 2 .\sqrt 2 + 5\sqrt 2 .\sqrt 2 - 2\sqrt 6 .\sqrt 6 \\
= 2 + 10 - 12 = 0\\
b.\sqrt {6 + 2\sqrt 5 } = \sqrt {5 + 2.\sqrt 5 .1 + 1} \\
= \sqrt {{{\left( {\sqrt 5 + 1} \right)}^2}} = \sqrt 5 + 1\\
4A.b.\dfrac{{\sqrt {36 - 12\sqrt 5 } }}{{\sqrt 6 }} = \dfrac{{\sqrt {6\left( {6 - 2\sqrt 5 } \right)} }}{{\sqrt 6 }}\\
= \sqrt {6 - 2\sqrt 5 } = \sqrt {5 - 2.\sqrt 5 .1 + 1} \\
= \sqrt {{{\left( {\sqrt 5 - 1} \right)}^2}} = \sqrt 5 - 1\\
6A.a.\sqrt {\dfrac{{ - 2t.\left( { - 3t} \right)}}{{3.8}}} = \sqrt {\dfrac{{6{t^2}}}{{24}}} \\
= \left| t \right|.\sqrt {\dfrac{1}{4}} \\
= \dfrac{{ - t}}{2}\left( {Do:t \le 0} \right)\\
b.\sqrt {x - \sqrt {{x^2} - 1} } .\sqrt {x + \sqrt {{x^2} - 1} } \\
= \sqrt {{x^2} - {{\left( {\sqrt {{x^2} - 1} } \right)}^2}} \\
= \sqrt {{x^2} - {x^2} + 1} = 1
\end{array}\)
