Rút gọn biểu thức \(B = \frac{{\left( {a\sqrt b + b} \right)\left( {\sqrt a + \sqrt b } \right)}}{{a - b}}.\sqrt {\frac{{ab + {b^2} - 2\sqrt {a{b^3}} }}{{a\left( {a + 2\sqrt b } \right) + b}}} \)

với (a, b > 0) được

Đáp án đúng là: B

\(B = \frac{{\left( {a\sqrt b + b} \right)\left( {\sqrt a + \sqrt b } \right)}}{{a - b}}.\sqrt {\frac{{ab + {b^2} - 2\sqrt {a{b^3}} }}{{a\left( {a + 2\sqrt b } \right) + b}}} \)

\(B = \frac{{\left( {a\sqrt b + b} \right)\left( {\sqrt a + \sqrt b } \right)}}{{a - b}}.\sqrt {\frac{{ab + {b^2} - 2\sqrt {a{b^3}} }}{{{a^2} + 2a\sqrt b + b}}} \)

\(B = \frac{{\left( {a\sqrt b + b} \right)\left( {\sqrt a + \sqrt b } \right)}}{{a - b}}.\sqrt {\frac{{{{\left( {\sqrt {ab} - b} \right)}^2}}}{{{{\left( {a + \sqrt b } \right)}^2}}}} \)

\(B = \frac{{\left( {a\sqrt b + b} \right)\left( {\sqrt a + \sqrt b } \right)}}{{a - b}}.\frac{{{{\left( {\sqrt {ab} - b} \right)}^2}}}{{{{\left( {a + \sqrt b } \right)}^2}}}\)

\(B = \frac{{\left( {a\sqrt b + b} \right)\left( {\sqrt a + \sqrt b } \right)}}{{a - b}}.\frac{{\left( {\sqrt {ab} - b} \right)}}{{\left( {a + \sqrt b } \right)}}\)

\(B = \frac{{\left( {a\sqrt b + b} \right)}}{{\left( {\sqrt a - \sqrt b } \right)}}.\frac{{\left( {\sqrt {ab} - b} \right)}}{{\left( {a + \sqrt b } \right)}} = \frac{{\sqrt b .\sqrt b .\left( {\sqrt a - \sqrt b } \right)}}{{\sqrt a - \sqrt b }} = b\).