Đáp án:
\(\begin{array}{l}
B13:\\
a)a > 0;b > 0\\
b)P = a - b\\
c)P = \sqrt 3 \\
B14:\\
a)\dfrac{{\sqrt a - 2}}{{3\sqrt a }}\\
b)a > 4
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
B13:\\
a)DK:a > 0;b > 0\\
b)P = \dfrac{{a - 2\sqrt {ab} + b + 4\sqrt {ab} }}{{\sqrt a + \sqrt b }}.\dfrac{{\sqrt {ab} \left( {\sqrt a - \sqrt b } \right)}}{{\sqrt {ab} }}\\
= \dfrac{{a + 2\sqrt {ab} + b}}{{\sqrt a + \sqrt b }}.\dfrac{{\sqrt {ab} \left( {\sqrt a - \sqrt b } \right)}}{{\sqrt {ab} }}\\
= \dfrac{{{{\left( {\sqrt a + \sqrt b } \right)}^2}}}{{\sqrt a + \sqrt b }}.\left( {\sqrt a - \sqrt b } \right)\\
= \left( {\sqrt a + \sqrt b } \right)\left( {\sqrt a - \sqrt b } \right)\\
= a - b\\
c)Thay:a = 2\sqrt 3 ;b = \sqrt 3 \\
\to P = 2\sqrt 3 - \sqrt 3 = \sqrt 3 \\
B14:\\
a)DK:a > 0;a \ne \left\{ {1;4} \right\}\\
Q = \dfrac{{\sqrt a - \sqrt a + 1}}{{\sqrt a \left( {\sqrt a - 1} \right)}}:\dfrac{{\left( {\sqrt a + 1} \right)\left( {\sqrt a - 1} \right) - \left( {\sqrt a + 2} \right)\left( {\sqrt a - 2} \right)}}{{\left( {\sqrt a - 2} \right)\left( {\sqrt a - 1} \right)}}\\
= \dfrac{1}{{\sqrt a \left( {\sqrt a - 1} \right)}}.\dfrac{{\left( {\sqrt a - 2} \right)\left( {\sqrt a - 1} \right)}}{{a - 1 - a + 4}}\\
= \dfrac{1}{{\sqrt a \left( {\sqrt a - 1} \right)}}.\dfrac{{\left( {\sqrt a - 2} \right)\left( {\sqrt a - 1} \right)}}{3}\\
= \dfrac{{\sqrt a - 2}}{{3\sqrt a }}\\
b)Q > 0\\
\to \dfrac{{\sqrt a - 2}}{{3\sqrt a }} > 0\\
\to \sqrt a - 2 > 0\left( {do:\sqrt a > 0\forall a > 0} \right)\\
\to a > 4
\end{array}\)
