Đáp án:
\(\dfrac{{a\sqrt a + 3a\sqrt b - 5b\sqrt a + b\sqrt b }}{{\sqrt a + \sqrt b }}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:a > 0;b > 0\\
P = {\left( {\sqrt a - \sqrt b } \right)^2} + \dfrac{{4\sqrt {ab} }}{{\sqrt a + \sqrt b }}.\dfrac{{a\sqrt b - b\sqrt a }}{{\sqrt {ab} }}\\
= a - 2\sqrt {ab} + b + \dfrac{4}{{\sqrt a + \sqrt b }}.\dfrac{{a\sqrt b - b\sqrt a }}{1}\\
= a - 2\sqrt {ab} + b + \dfrac{{4a\sqrt b - 4b\sqrt a }}{{\sqrt a + \sqrt b }}\\
= \dfrac{{\left( {a - 2\sqrt {ab} + b} \right)\left( {\sqrt a + \sqrt b } \right) + 4a\sqrt b - 4b\sqrt a }}{{\sqrt a + \sqrt b }}\\
= \dfrac{{a\sqrt a + a\sqrt b - 2a\sqrt b - 2b\sqrt a + b\sqrt a + b\sqrt b + 4a\sqrt b - 4b\sqrt a }}{{\sqrt a + \sqrt b }}\\
= \dfrac{{a\sqrt a + 3a\sqrt b - 5b\sqrt a + b\sqrt b }}{{\sqrt a + \sqrt b }}\\
b.Thay:a = 2\sqrt 3 ;b = \sqrt 3 \\
\to P = \dfrac{{2\sqrt 3 \sqrt {2\sqrt 3 } + 3.2\sqrt 3 \sqrt {\sqrt 3 } - 5.\sqrt 3 \sqrt {2\sqrt 3 } + \sqrt 3 \sqrt {\sqrt 3 } }}{{\sqrt {2\sqrt 3 } + \sqrt {\sqrt 3 } }}\\
= \dfrac{{ - 3\sqrt 3 .\sqrt {2\sqrt 3 } + 7\sqrt 3 \sqrt {\sqrt 3 } }}{{\sqrt {2\sqrt 3 } + \sqrt {\sqrt 3 } }}
\end{array}\)
