Đáp án:
c) 1>a>0
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:1 > a > - 1\\
B = \dfrac{{3 + \sqrt {\left( {1 - a} \right)\left( {1 + a} \right)} }}{{\sqrt {1 + a} }}:\dfrac{{3 + \sqrt {\left( {1 - a} \right)\left( {1 + a} \right)} }}{{\sqrt {\left( {1 - a} \right)\left( {1 + a} \right)} }}\\
= \dfrac{{3 + \sqrt {\left( {1 - a} \right)\left( {1 + a} \right)} }}{{\sqrt {1 + a} }}.\dfrac{{\sqrt {\left( {1 - a} \right)\left( {1 + a} \right)} }}{{3 + \sqrt {\left( {1 - a} \right)\left( {1 + a} \right)} }}\\
= \sqrt {1 - a} \\
b)Thay:a = \dfrac{{\sqrt 3 }}{{2 + \sqrt 3 }}\\
\to B = \sqrt {1 - \dfrac{{\sqrt 3 }}{{2 + \sqrt 3 }}} = \sqrt {\dfrac{{2 + \sqrt 3 - \sqrt 3 }}{{2 + \sqrt 3 }}} \\
= \sqrt {\dfrac{2}{{2 + \sqrt 3 }}} = \sqrt {\dfrac{4}{{4 + 2\sqrt 3 }}} = \dfrac{2}{{\sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} }}\\
= \dfrac{2}{{\sqrt 3 + 1}} = - 1 + \sqrt 3 \\
c)\sqrt B > B\\
\to B - \sqrt B < 0\\
\to \sqrt B \left( {\sqrt B - 1} \right) < 0\\
\to \sqrt B - 1 < 0\left( {do:\sqrt B \ge 0\forall B} \right)\\
\to B < 1\\
\to \sqrt {1 - a} < 1\\
\to 1 - a < 1\\
\to a > 0\\
KL:1 > a > 0
\end{array}\)
