Đáp án:
i. 5
k. 1
Giải thích các bước giải:
\(\begin{array}{l}
i.\dfrac{{\sqrt 7 \left( {\sqrt 3 - 7} \right)}}{{ - \left( {\sqrt 3 - 7} \right)}} - \dfrac{{18}}{{\sqrt 7 - 5}} = - \sqrt 7 - \dfrac{{18}}{{\sqrt 7 - 5}}\\
= \dfrac{{ - 7 + 5\sqrt 7 - 18}}{{\sqrt 7 - 5}} = \dfrac{{ - 25 + 5\sqrt 7 }}{{\sqrt 7 - 5}}\\
= \dfrac{{5\left( { - 5 + \sqrt 7 } \right)}}{{\sqrt 7 - 5}} = 5\\
k.\dfrac{5}{{\sqrt 7 + \sqrt 2 }} + \dfrac{{\sqrt 2 + 1}}{{2 - 1}} - \dfrac{{\sqrt 7 \left( {\sqrt 7 + 1} \right)}}{{\sqrt 7 + 1}}\\
= \dfrac{5}{{\sqrt 7 + \sqrt 2 }} + \sqrt 2 + 1 - \sqrt 7 \\
= \dfrac{{5 + \sqrt {14} + 2 + \sqrt 7 + \sqrt 2 - 7 - \sqrt {14} }}{{\sqrt 7 + \sqrt 2 }}\\
= \dfrac{{\sqrt 7 + \sqrt 2 }}{{\sqrt 7 + \sqrt 2 }} = 1\\
o.\sqrt {3 - 2\sqrt 3 .1 + 1} + \sqrt {\dfrac{{2\left( {2 + \sqrt 3 } \right)}}{{4 - 3}}} - 3\sqrt 3 \\
= \sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} + \sqrt {4 + 2\sqrt 3 } - 3\sqrt 3 \\
= \sqrt 3 - 1 + \sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} - 3\sqrt 3 \\
= - 2\sqrt 3 - 1 + \sqrt 3 + 1\\
= - \sqrt 3 \\
q.\sqrt {\dfrac{{8\left( {8 - 3\sqrt 7 } \right)}}{{64 - {{\left( {3\sqrt 7 } \right)}^2}}}} + \sqrt {\dfrac{{\left( {38 - 14\sqrt 7 } \right)\left( {3 + \sqrt 7 } \right)}}{{9 - 7}}} \\
= \sqrt {64 - 24\sqrt 7 } + \sqrt {\dfrac{{114 - 42\sqrt 7 + 38\sqrt 7 - 98}}{2}} \\
= \sqrt {64 - 24\sqrt 7 } + \dfrac{{\sqrt {16 - 4\sqrt 7 } }}{{\sqrt 2 }}\\
= \sqrt {64 - 24\sqrt 7 } + \sqrt {8 - 2\sqrt 7 } \\
= \sqrt {36 - 2.6.2\sqrt 7 + {{\left( {2\sqrt 7 } \right)}^2}} + \sqrt {7 - 2.\sqrt 7 .1 + 1} \\
= \sqrt {{{\left( {6 - 2\sqrt 7 } \right)}^2}} + \sqrt {{{\left( {\sqrt 7 - 1} \right)}^2}} \\
= 6 - 2\sqrt 7 + \sqrt 7 - 1 = 5 - \sqrt 7
\end{array}\)
