Đáp án:
B=1
Giải thích các bước giải:
\(\begin{array}{l}
x = \dfrac{1}{2}\left[ {\dfrac{{\sqrt {1 - a} .\sqrt {1 - a} - \sqrt a .\sqrt a }}{{\sqrt a .\sqrt {\left( {1 - a} \right)} }}} \right]\\
= \dfrac{1}{2}.\dfrac{{1 - a - a}}{{\sqrt {a\left( {1 - a} \right)} }}\\
= \dfrac{{1 - 2a}}{{2\sqrt {a\left( {1 - a} \right)} }}\\
\to {x^2} = \dfrac{{1 - 4a + 4{a^2}}}{{4\left( {a\left( {1 - a} \right)} \right)}} = \dfrac{{1 - 4a + 4{a^2}}}{{4a\left( {1 - a} \right)}}\\
= \dfrac{{1 - 4a + 4{a^2}}}{{4a - 4{a^2}}}\\
B = \dfrac{{2a\sqrt {1 + \dfrac{{1 - 4a + 4{a^2}}}{{4a - 4{a^2}}}} }}{{\sqrt {1 + \dfrac{{1 - 4a + 4{a^2}}}{{4a - 4{a^2}}}} - \dfrac{{1 - 2a}}{{2\sqrt {a\left( {1 - a} \right)} }}}}\\
= \dfrac{{2a.\sqrt {\dfrac{{4a - 4{a^2} + 1 - 4a + 4{a^2}}}{{4a - 4{a^2}}}} }}{{\sqrt {\dfrac{{4a - 4{a^2} + 1 - 4a + 4{a^2}}}{{4a - 4{a^2}}}} - \dfrac{{1 - 2a}}{{2\sqrt {a\left( {1 - a} \right)} }}}}\\
= \dfrac{{2a\sqrt {\dfrac{1}{{4a - 4{a^2}}}} }}{{\sqrt {\dfrac{1}{{4a - 4{a^2}}}} - \dfrac{{1 - 2a}}{{2\sqrt {a\left( {1 - a} \right)} }}}}\\
= \dfrac{a}{{\sqrt {a - {a^2}} }}:\left[ {\dfrac{1}{{2\sqrt {a - {a^2}} }} - \dfrac{{1 - 2a}}{{2\sqrt {a\left( {1 - a} \right)} }}} \right]\\
= \dfrac{{\sqrt a }}{{\sqrt {1 - a} }}:\left[ {\dfrac{{1 - 1 + 2a}}{{2\sqrt {a\left( {1 - a} \right)} }}} \right]\\
= \dfrac{{\sqrt a }}{{\sqrt {1 - a} }}.\dfrac{{2\sqrt {a\left( {1 - a} \right)} }}{{2a}}\\
= \dfrac{a}{a} = 1
\end{array}\)
