Đáp án:
b) B=1
Giải thích các bước giải:
\(\begin{array}{l}
a)A = \dfrac{{\sqrt {2 + \sqrt 3 } }}{{\sqrt {2 - \sqrt 3 } }} + \dfrac{{\sqrt {2 - \sqrt 3 } }}{{\sqrt {2 + \sqrt 3 } }}\\
= \dfrac{{\sqrt {4 + 2\sqrt 3 } }}{{\sqrt {4 - 2\sqrt 3 } }} + \dfrac{{\sqrt {4 - 2\sqrt 3 } }}{{\sqrt {4 + 2\sqrt 3 } }}\\
= \dfrac{{\sqrt {3 + 2\sqrt 3 .1 + 1} }}{{\sqrt {3 - 2\sqrt 3 .1 + 1} }} + \dfrac{{\sqrt {3 - 2\sqrt 3 .1 + 1} }}{{\sqrt {3 + 2\sqrt 3 .1 + 1} }}\\
= \dfrac{{\sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} }}{{\sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} }} + \dfrac{{\sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} }}{{\sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} }}\\
= \dfrac{{\sqrt 3 + 1}}{{\sqrt 3 - 1}} + \dfrac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}}\\
= \dfrac{{4 + 2\sqrt 3 + 4 - 2\sqrt 3 }}{{3 - 1}}\\
= \dfrac{8}{2} = 4\\
b)B = \left[ {\dfrac{{\left( {1 - \sqrt a } \right)\left( {a + \sqrt a + 1} \right)}}{{1 - \sqrt a }} + \sqrt a } \right].{\left[ {\dfrac{{1 - \sqrt a }}{{\left( {1 - \sqrt a } \right)\left( {1 + \sqrt a } \right)}}} \right]^2}\\
= \left( {a + \sqrt a + 1 + \sqrt a } \right).\dfrac{1}{{{{\left( {1 + \sqrt a } \right)}^2}}}\\
= {\left( {1 + \sqrt a } \right)^2}.\dfrac{1}{{{{\left( {1 + \sqrt a } \right)}^2}}} = 1\\
c)C = \dfrac{{1 + \sqrt a }}{{\sqrt a \left( {\sqrt a - 1} \right)}}.\dfrac{{{{\left( {\sqrt a - 1} \right)}^2}}}{{\sqrt a + 1}}\\
= \dfrac{{\sqrt a - 1}}{{\sqrt a }}\\
d)D = \sqrt {x - 2 - 2.\sqrt {x - 2} .1 + 1} \\
= \sqrt {{{\left( {\sqrt {x - 2} - 1} \right)}^2}} \\
= \left| {\sqrt {x - 2} - 1} \right|\\
= \sqrt {x - 2} - 1\left( {do:x \ge 2} \right)
\end{array}\)
