Cho biểu thức: P = ((x - y) / (căn bậc hai x - căn bậc hai y) + (căn bậc hai x^3
Câu hỏi:
Cho biểu thức: \(P = \left( {\frac{{x - y}}{{\sqrt x - \sqrt y }} + \frac{{\sqrt {{x^3}} - \sqrt {{y^3}} }}{{y - x}}} \right):\frac{{{{\left( {\sqrt x - \sqrt y } \right)}^2} + \sqrt {xy} }}{{\sqrt x + \sqrt y }}\) với x ≥ 0, y ≥ 0, x ≠ y.
a) Rút gọn A.
b) Chứng minh rằng A ≥ 0.
Trả lời:
a) Với x ≥ 0, y ≥ 0, x ≠ y. Ta có:
\(P = \left( {\frac{{x - y}}{{\sqrt x - \sqrt y }} + \frac{{\sqrt {{x^3}} - \sqrt {{y^3}} }}{{y - x}}} \right):\frac{{{{\left( {\sqrt x - \sqrt y } \right)}^2} + \sqrt {xy} }}{{\sqrt x + \sqrt y }}\)
\(P = \left( {\frac{{\left( {\sqrt x - \sqrt y } \right)\left( {\sqrt x + \sqrt y } \right)}}{{\sqrt x - \sqrt y }} + \frac{{\left( {\sqrt x - \sqrt y } \right)\left( {x + \sqrt {xy} + y} \right)}}{{y - x}}} \right):\frac{{{{\left( {\sqrt x - \sqrt y } \right)}^2} + \sqrt {xy} }}{{\sqrt x + \sqrt y }}\)
\(P = \left( {\sqrt x + \sqrt y - \frac{{\left( {\sqrt x - \sqrt y } \right)\left( {x + \sqrt {xy} + y} \right)}}{{\left( {\sqrt x - \sqrt y } \right)\left( {\sqrt x + \sqrt y } \right)}}} \right):\frac{{{{\left( {\sqrt x - \sqrt y } \right)}^2} + \sqrt {xy} }}{{\sqrt x + \sqrt y }}\)
\(P = \left( {\sqrt x + \sqrt y - \frac{{x + \sqrt {xy} + y}}{{\sqrt x + \sqrt y }}} \right):\frac{{{{\left( {\sqrt x - \sqrt y } \right)}^2} + \sqrt {xy} }}{{\sqrt x + \sqrt y }}\)
\[P = \frac{{{{\left( {\sqrt x + \sqrt y } \right)}^2} - x - \sqrt {xy} - y}}{{\sqrt x + \sqrt y }}.\frac{{\sqrt x + \sqrt y }}{{x - \sqrt {xy} + y}}\]
\[P = \frac{{\sqrt {xy} }}{{x - \sqrt {xy} + y}}\].
b) Ta có: x ≥ 0, y ≥ 0, x ≠ y thì \(\sqrt {xy} \ge 0\)
\[x - \sqrt {xy} + y = {\left( {\sqrt x - \sqrt y } \right)^2} + \sqrt {xy} \ge 0\]
Vậy A ≥ 0 với x ≥ 0, y ≥ 0, x ≠ y.
