Đáp án:
\(\begin{array}{l}
R = \dfrac{{3\sqrt 3 - 1}}{2}\\
T = 26\\
W = 9\\
II = 4
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
R = \sqrt {\dfrac{{13 + 4\sqrt 3 }}{4}} - \sqrt {\dfrac{{7 - 4\sqrt 3 }}{4}} \\
= \sqrt {\dfrac{{12 + 2.2\sqrt 3 .1 + 1}}{4}} - \sqrt {\dfrac{{4 - 2.2.\sqrt 3 + 3}}{4}} \\
= \sqrt {\dfrac{{{{\left( {2\sqrt 3 + 1} \right)}^2}}}{4}} - \sqrt {\dfrac{{{{\left( {2 - \sqrt 3 } \right)}^2}}}{4}} \\
= \dfrac{{2\sqrt 3 + 1}}{2} - \dfrac{{2 - \sqrt 3 }}{2}\\
= \dfrac{{2\sqrt 3 + 1 - 2 + \sqrt 3 }}{2}\\
= \dfrac{{3\sqrt 3 - 1}}{2}\\
T = \sqrt {225 - 2.15.\sqrt 2 + 2} + \sqrt {121 + 2.11.\sqrt 2 + 2} \\
= \sqrt {{{\left( {15 - \sqrt 2 } \right)}^2}} + \sqrt {{{\left( {11 + \sqrt 2 } \right)}^2}} \\
= 15 - \sqrt 2 + 11 + \sqrt 2 \\
= 26\\
W = \sqrt {28 + 10\sqrt 3 } + \sqrt {19 - 8\sqrt 3 } \\
= \sqrt {25 + 2.5.\sqrt 3 + 3} + \sqrt {16 - 2.4.\sqrt 3 + 3} \\
= \sqrt {{{\left( {5 + \sqrt 3 } \right)}^2}} + \sqrt {{{\left( {4 - \sqrt 3 } \right)}^2}} \\
= 5 + \sqrt 3 + 4 - \sqrt 3 \\
= 9\\
II = \sqrt {16 + 2.4.\sqrt 7 + 7} - \sqrt 7 \\
= \sqrt {{{\left( {4 + \sqrt 7 } \right)}^2}} - \sqrt 7 \\
= 4 + \sqrt 7 - \sqrt 7 = 4
\end{array}\)
