Rút gọn \(\dfrac{{{{\left( {\sqrt a - \sqrt b } \right)}^3} + 2a\sqrt a + b\sqrt b }}{{a\sqrt a + b\sqrt b }} + \dfrac{{3\left( {\sqrt {ab} - b} \right)}}{{a - b}}\) với \(a > 0,\,...

Câu hỏi :

Rút gọn \(\dfrac{{{{\left( {\sqrt a - \sqrt b } \right)}^3} + 2a\sqrt a + b\sqrt b }}{{a\sqrt a + b\sqrt b }} + \dfrac{{3\left( {\sqrt {ab} - b} \right)}}{{a - b}}\) với \(a > 0,\,\,b > 0,\,\,a \ne b\)

A. 1

B. 2

C. 3

D. 4

* Đáp án

C

* Hướng dẫn giải

ĐKXĐ : \(a > 0,\,\,b > 0,\,\,a \ne b\)

\(\dfrac{{{{\left( {\sqrt a - \sqrt b } \right)}^3} + 2a\sqrt a + b\sqrt b }}{{a\sqrt a + b\sqrt b }} + \dfrac{{3\left( {\sqrt {ab} - b} \right)}}{{a - b}}\)

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

\( = \dfrac{{3a\sqrt a  - 3a\sqrt b  + 3b\sqrt a }}{{\left( {\sqrt a  + \sqrt b } \right)\left( {a - \sqrt {ab}  + b} \right)}} \)\(+ \dfrac{{3\sqrt b \left( {\sqrt a  - \sqrt b } \right)}}{{\left( {\sqrt a  + \sqrt b } \right)\left( {\sqrt a  - \sqrt b } \right)}}\)

\( = \dfrac{{3\sqrt a \left( {a - \sqrt {ab}  + b} \right)}}{{\left( {\sqrt a  + \sqrt b } \right)\left( {a - \sqrt {ab}  + b} \right)}}\)\( + \dfrac{{3\sqrt b }}{{\sqrt a  + \sqrt b }}\)

\( = \dfrac{{3\sqrt a }}{{\sqrt a  + \sqrt b }} + \dfrac{{3\sqrt b }}{{\sqrt a  + \sqrt b }}\)

\( = \dfrac{{3\left( {\sqrt a  + \sqrt b } \right)}}{{\sqrt a  + \sqrt b }} = 3 .\)

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