Cho biểu thức:  (A = left( {frac{{x - y}}{{sqrt x  - sqrt y }} + frac{{sqrt {{x^3}}  - sqrt {{y^3}} }}{{y - x}}} ight):

a) ĐKXĐ: \(x \ge 0;y \ge 0;x,y\) không đồng thời bằng 0

\(A = \left( {\frac{{x - y}}{{\sqrt x  - \sqrt y }} + \frac{{\sqrt {{x^3}}  - \sqrt {{y^3}} }}{{y - x}}} \right):\frac{{{{\left( {\sqrt x  - \sqrt y } \right)}^2} + \sqrt {xy} }}{{\sqrt x  + \sqrt y }}\)

\(\begin{array}{l}
 = \left( {\frac{{x - y}}{{\sqrt x  - \sqrt y }} + \frac{{\sqrt {{x^3}}  - \sqrt {{y^3}} }}{{y - x}}} \right):\frac{{{{\left( {\sqrt x  - \sqrt y } \right)}^2} + \sqrt {xy} }}{{\sqrt x  + \sqrt y }}\\
 = \left[ {\frac{{\left( {\sqrt x  - \sqrt y } \right)\left( {\sqrt x  + \sqrt y } \right)}}{{\left( {\sqrt x  - \sqrt y } \right)}} + \frac{{\left( {\sqrt x  - \sqrt y } \right)\left( {x + \sqrt {xy}  + y} \right)}}{{\left( {\sqrt y  - \sqrt x } \right)\left( {\sqrt y  + \sqrt x } \right)}}} \right]:\frac{{x - 2\sqrt {xy}  + y + \sqrt {xy} }}{{\sqrt x  + \sqrt y }}\\
 = \left( {\sqrt x  + \sqrt y  + \frac{{x + \sqrt {xy}  + y}}{{\sqrt x  + \sqrt y }}} \right).\frac{{\sqrt x  + \sqrt y }}{{x - \sqrt {xy}  + y}}\\
 = \frac{{x + 2\sqrt {xy}  + y - x - \sqrt {xy}  - y}}{{x - \sqrt {xy}  + \sqrt y }} = \frac{{\sqrt {xy} }}{{x - \sqrt {xy}  + y}}
\end{array}\)

b) Vì \(x \ge 0;y \ge 0 \Rightarrow \sqrt {xy}  \ge 0.\)

Ta có: \(x - \sqrt {xy}  + y = {\left( {\sqrt x } \right)^2} - 2.\frac{1}{2}.\sqrt x .\sqrt y  + {\left( {\frac{1}{2}\sqrt y } \right)^2} + y - {\left( {\frac{1}{2}\sqrt y } \right)^2}\)

\( = {\left( {\sqrt x  - \frac{1}{2}\sqrt y } \right)^2} + \frac{{3y}}{4} \ge 0\) (vì \( = {\left( {\sqrt x  - \frac{1}{2}\sqrt y } \right)^2} \ge 0\) và \(y \ge 0\) nên \(\frac{{3y}}{4} \ge 0\)

Do đó: \(x - \sqrt {xy}  + y \ge 0\)

Vậy \(A = \frac{{\sqrt {xy} }}{{x - \sqrt {xy}  + y}} \ge 0\) (với mọi \(x \ge 0;y \ge 0;x,y\) không đồng thời bằng 0)