a) Điều kiện \(x \geq 0\), \(y \geq 0\), \(x \neq y\).

Khi đó ta có:

\(P = \left(\right. \frac{2 \sqrt{x y}}{x - y} - \frac{\sqrt{x} + \sqrt{y}}{2 \sqrt{x} - 2 \sqrt{y}} \left.\right) . \frac{2 \sqrt{x}}{\sqrt{x} - \sqrt{y}}\)

\(= \frac{4 \sqrt{x y} - \left(\right. \sqrt{x} + \sqrt{y} \left.\right)^{2}}{2 \left(\right. \sqrt{x} - \sqrt{y} \left.\right) \left(\right. \sqrt{x} + \sqrt{y} \left.\right)} . \frac{2 \sqrt{x}}{\sqrt{x} - \sqrt{y}}\)

\(= \frac{4 \sqrt{x y} - x - 2 \sqrt{x y} - y}{2 \left(\right. \sqrt{x} - \sqrt{y} \left.\right) \left(\right. \sqrt{x} + \sqrt{y} \left.\right)} . \frac{2 \sqrt{x}}{\sqrt{x} - \sqrt{y}}\)

\(= \frac{- \left(\right. \sqrt{x} - \sqrt{y} \left.\right)^{2}}{2 \left(\right. \sqrt{x} - \sqrt{y} \left.\right) \left(\right. \sqrt{x} + \sqrt{y} \left.\right)} . \frac{2 \sqrt{x}}{\sqrt{x} - \sqrt{y}}\)

\(= \frac{- \sqrt{x}}{\sqrt{x} + \sqrt{y}}\)

b) Ta có \(\frac{x}{y} = \frac{4}{9}\) suy ra \(y = \frac{9}{4} x\)

Do đó \(P = \frac{- \sqrt{x}}{\sqrt{x} + \sqrt{\frac{9 x}{4}}}\)

\(= \frac{- \sqrt{x}}{\sqrt{x} + \frac{3}{2} \sqrt{x}}\)

\(= \frac{- \sqrt{x}}{\frac{5}{2} \sqrt{x}} = - \frac{2}{5}\).

a) Điều kiện \(x \geq 0\), \(y \geq 0\), \(x \neq y\).

Khi đó ta có:

\(P = \left(\right. \frac{2 \sqrt{x y}}{x - y} - \frac{\sqrt{x} + \sqrt{y}}{2 \sqrt{x} - 2 \sqrt{y}} \left.\right) . \frac{2 \sqrt{x}}{\sqrt{x} - \sqrt{y}}\)

\(= \frac{4 \sqrt{x y} - \left(\right. \sqrt{x} + \sqrt{y} \left.\right)^{2}}{2 \left(\right. \sqrt{x} - \sqrt{y} \left.\right) \left(\right. \sqrt{x} + \sqrt{y} \left.\right)} . \frac{2 \sqrt{x}}{\sqrt{x} - \sqrt{y}}\)

\(= \frac{4 \sqrt{x y} - x - 2 \sqrt{x y} - y}{2 \left(\right. \sqrt{x} - \sqrt{y} \left.\right) \left(\right. \sqrt{x} + \sqrt{y} \left.\right)} . \frac{2 \sqrt{x}}{\sqrt{x} - \sqrt{y}}\)

\(= \frac{- \left(\right. \sqrt{x} - \sqrt{y} \left.\right)^{2}}{2 \left(\right. \sqrt{x} - \sqrt{y} \left.\right) \left(\right. \sqrt{x} + \sqrt{y} \left.\right)} . \frac{2 \sqrt{x}}{\sqrt{x} - \sqrt{y}}\)

\(= \frac{- \sqrt{x}}{\sqrt{x} + \sqrt{y}}\)

b) Ta có \(\frac{x}{y} = \frac{4}{9}\) suy ra \(y = \frac{9}{4} x\)

Do đó \(P = \frac{- \sqrt{x}}{\sqrt{x} + \sqrt{\frac{9 x}{4}}}\)

\(= \frac{- \sqrt{x}}{\sqrt{x} + \frac{3}{2} \sqrt{x}}\)

\(= \frac{- \sqrt{x}}{\frac{5}{2} \sqrt{x}} = - \frac{2}{5}\).

a) Điều kiện \(x \geq 0\), \(y \geq 0\), \(x \neq y\).

Khi đó ta có:

\(P = \left(\right. \frac{2 \sqrt{x y}}{x - y} - \frac{\sqrt{x} + \sqrt{y}}{2 \sqrt{x} - 2 \sqrt{y}} \left.\right) . \frac{2 \sqrt{x}}{\sqrt{x} - \sqrt{y}}\)

\(= \frac{4 \sqrt{x y} - \left(\right. \sqrt{x} + \sqrt{y} \left.\right)^{2}}{2 \left(\right. \sqrt{x} - \sqrt{y} \left.\right) \left(\right. \sqrt{x} + \sqrt{y} \left.\right)} . \frac{2 \sqrt{x}}{\sqrt{x} - \sqrt{y}}\)

\(= \frac{4 \sqrt{x y} - x - 2 \sqrt{x y} - y}{2 \left(\right. \sqrt{x} - \sqrt{y} \left.\right) \left(\right. \sqrt{x} + \sqrt{y} \left.\right)} . \frac{2 \sqrt{x}}{\sqrt{x} - \sqrt{y}}\)

\(= \frac{- \left(\right. \sqrt{x} - \sqrt{y} \left.\right)^{2}}{2 \left(\right. \sqrt{x} - \sqrt{y} \left.\right) \left(\right. \sqrt{x} + \sqrt{y} \left.\right)} . \frac{2 \sqrt{x}}{\sqrt{x} - \sqrt{y}}\)

\(= \frac{- \sqrt{x}}{\sqrt{x} + \sqrt{y}}\)

b) Ta có \(\frac{x}{y} = \frac{4}{9}\) suy ra \(y = \frac{9}{4} x\)

Do đó \(P = \frac{- \sqrt{x}}{\sqrt{x} + \sqrt{\frac{9 x}{4}}}\)

\(= \frac{- \sqrt{x}}{\sqrt{x} + \frac{3}{2} \sqrt{x}}\)

\(= \frac{- \sqrt{x}}{\frac{5}{2} \sqrt{x}} = - \frac{2}{5}\).

a) Điều kiện \(x \geq 0\), \(y \geq 0\), \(x \neq y\).

Khi đó ta có:

\(P = \left(\right. \frac{2 \sqrt{x y}}{x - y} - \frac{\sqrt{x} + \sqrt{y}}{2 \sqrt{x} - 2 \sqrt{y}} \left.\right) . \frac{2 \sqrt{x}}{\sqrt{x} - \sqrt{y}}\)

\(= \frac{4 \sqrt{x y} - \left(\right. \sqrt{x} + \sqrt{y} \left.\right)^{2}}{2 \left(\right. \sqrt{x} - \sqrt{y} \left.\right) \left(\right. \sqrt{x} + \sqrt{y} \left.\right)} . \frac{2 \sqrt{x}}{\sqrt{x} - \sqrt{y}}\)

\(= \frac{4 \sqrt{x y} - x - 2 \sqrt{x y} - y}{2 \left(\right. \sqrt{x} - \sqrt{y} \left.\right) \left(\right. \sqrt{x} + \sqrt{y} \left.\right)} . \frac{2 \sqrt{x}}{\sqrt{x} - \sqrt{y}}\)

\(= \frac{- \left(\right. \sqrt{x} - \sqrt{y} \left.\right)^{2}}{2 \left(\right. \sqrt{x} - \sqrt{y} \left.\right) \left(\right. \sqrt{x} + \sqrt{y} \left.\right)} . \frac{2 \sqrt{x}}{\sqrt{x} - \sqrt{y}}\)

\(= \frac{- \sqrt{x}}{\sqrt{x} + \sqrt{y}}\)

b) Ta có \(\frac{x}{y} = \frac{4}{9}\) suy ra \(y = \frac{9}{4} x\)

Do đó \(P = \frac{- \sqrt{x}}{\sqrt{x} + \sqrt{\frac{9 x}{4}}}\)

\(= \frac{- \sqrt{x}}{\sqrt{x} + \frac{3}{2} \sqrt{x}}\)

\(= \frac{- \sqrt{x}}{\frac{5}{2} \sqrt{x}} = - \frac{2}{5}\).

a) Điều kiện \(x \geq 0\), \(y \geq 0\), \(x \neq y\).

Khi đó ta có:

\(P = \left(\right. \frac{2 \sqrt{x y}}{x - y} - \frac{\sqrt{x} + \sqrt{y}}{2 \sqrt{x} - 2 \sqrt{y}} \left.\right) . \frac{2 \sqrt{x}}{\sqrt{x} - \sqrt{y}}\)

\(= \frac{4 \sqrt{x y} - \left(\right. \sqrt{x} + \sqrt{y} \left.\right)^{2}}{2 \left(\right. \sqrt{x} - \sqrt{y} \left.\right) \left(\right. \sqrt{x} + \sqrt{y} \left.\right)} . \frac{2 \sqrt{x}}{\sqrt{x} - \sqrt{y}}\)

\(= \frac{4 \sqrt{x y} - x - 2 \sqrt{x y} - y}{2 \left(\right. \sqrt{x} - \sqrt{y} \left.\right) \left(\right. \sqrt{x} + \sqrt{y} \left.\right)} . \frac{2 \sqrt{x}}{\sqrt{x} - \sqrt{y}}\)

\(= \frac{- \left(\right. \sqrt{x} - \sqrt{y} \left.\right)^{2}}{2 \left(\right. \sqrt{x} - \sqrt{y} \left.\right) \left(\right. \sqrt{x} + \sqrt{y} \left.\right)} . \frac{2 \sqrt{x}}{\sqrt{x} - \sqrt{y}}\)

\(= \frac{- \sqrt{x}}{\sqrt{x} + \sqrt{y}}\)

b) Ta có \(\frac{x}{y} = \frac{4}{9}\) suy ra \(y = \frac{9}{4} x\)

Do đó \(P = \frac{- \sqrt{x}}{\sqrt{x} + \sqrt{\frac{9 x}{4}}}\)

\(= \frac{- \sqrt{x}}{\sqrt{x} + \frac{3}{2} \sqrt{x}}\)

\(= \frac{- \sqrt{x}}{\frac{5}{2} \sqrt{x}} = - \frac{2}{5}\).

a) Điều kiện \(x \geq 0\), \(y \geq 0\), \(x \neq y\).

Khi đó ta có:

\(P = \left(\right. \frac{2 \sqrt{x y}}{x - y} - \frac{\sqrt{x} + \sqrt{y}}{2 \sqrt{x} - 2 \sqrt{y}} \left.\right) . \frac{2 \sqrt{x}}{\sqrt{x} - \sqrt{y}}\)

\(= \frac{4 \sqrt{x y} - \left(\right. \sqrt{x} + \sqrt{y} \left.\right)^{2}}{2 \left(\right. \sqrt{x} - \sqrt{y} \left.\right) \left(\right. \sqrt{x} + \sqrt{y} \left.\right)} . \frac{2 \sqrt{x}}{\sqrt{x} - \sqrt{y}}\)

\(= \frac{4 \sqrt{x y} - x - 2 \sqrt{x y} - y}{2 \left(\right. \sqrt{x} - \sqrt{y} \left.\right) \left(\right. \sqrt{x} + \sqrt{y} \left.\right)} . \frac{2 \sqrt{x}}{\sqrt{x} - \sqrt{y}}\)

\(= \frac{- \left(\right. \sqrt{x} - \sqrt{y} \left.\right)^{2}}{2 \left(\right. \sqrt{x} - \sqrt{y} \left.\right) \left(\right. \sqrt{x} + \sqrt{y} \left.\right)} . \frac{2 \sqrt{x}}{\sqrt{x} - \sqrt{y}}\)

\(= \frac{- \sqrt{x}}{\sqrt{x} + \sqrt{y}}\)

b) Ta có \(\frac{x}{y} = \frac{4}{9}\) suy ra \(y = \frac{9}{4} x\)

Do đó \(P = \frac{- \sqrt{x}}{\sqrt{x} + \sqrt{\frac{9 x}{4}}}\)

\(= \frac{- \sqrt{x}}{\sqrt{x} + \frac{3}{2} \sqrt{x}}\)

\(= \frac{- \sqrt{x}}{\frac{5}{2} \sqrt{x}} = - \frac{2}{5}\).

a) Điều kiện \(x \geq 0\), \(y \geq 0\), \(x \neq y\).

Khi đó ta có:

\(P = \left(\right. \frac{2 \sqrt{x y}}{x - y} - \frac{\sqrt{x} + \sqrt{y}}{2 \sqrt{x} - 2 \sqrt{y}} \left.\right) . \frac{2 \sqrt{x}}{\sqrt{x} - \sqrt{y}}\)

\(= \frac{4 \sqrt{x y} - \left(\right. \sqrt{x} + \sqrt{y} \left.\right)^{2}}{2 \left(\right. \sqrt{x} - \sqrt{y} \left.\right) \left(\right. \sqrt{x} + \sqrt{y} \left.\right)} . \frac{2 \sqrt{x}}{\sqrt{x} - \sqrt{y}}\)

\(= \frac{4 \sqrt{x y} - x - 2 \sqrt{x y} - y}{2 \left(\right. \sqrt{x} - \sqrt{y} \left.\right) \left(\right. \sqrt{x} + \sqrt{y} \left.\right)} . \frac{2 \sqrt{x}}{\sqrt{x} - \sqrt{y}}\)

\(= \frac{- \left(\right. \sqrt{x} - \sqrt{y} \left.\right)^{2}}{2 \left(\right. \sqrt{x} - \sqrt{y} \left.\right) \left(\right. \sqrt{x} + \sqrt{y} \left.\right)} . \frac{2 \sqrt{x}}{\sqrt{x} - \sqrt{y}}\)

\(= \frac{- \sqrt{x}}{\sqrt{x} + \sqrt{y}}\)

b) Ta có \(\frac{x}{y} = \frac{4}{9}\) suy ra \(y = \frac{9}{4} x\)

Do đó \(P = \frac{- \sqrt{x}}{\sqrt{x} + \sqrt{\frac{9 x}{4}}}\)

\(= \frac{- \sqrt{x}}{\sqrt{x} + \frac{3}{2} \sqrt{x}}\)

\(= \frac{- \sqrt{x}}{\frac{5}{2} \sqrt{x}} = - \frac{2}{5}\).

a) Điều kiện \(x \geq 0\), \(y \geq 0\), \(x \neq y\).

Khi đó ta có:

\(P = \left(\right. \frac{2 \sqrt{x y}}{x - y} - \frac{\sqrt{x} + \sqrt{y}}{2 \sqrt{x} - 2 \sqrt{y}} \left.\right) . \frac{2 \sqrt{x}}{\sqrt{x} - \sqrt{y}}\)

\(= \frac{4 \sqrt{x y} - \left(\right. \sqrt{x} + \sqrt{y} \left.\right)^{2}}{2 \left(\right. \sqrt{x} - \sqrt{y} \left.\right) \left(\right. \sqrt{x} + \sqrt{y} \left.\right)} . \frac{2 \sqrt{x}}{\sqrt{x} - \sqrt{y}}\)

\(= \frac{4 \sqrt{x y} - x - 2 \sqrt{x y} - y}{2 \left(\right. \sqrt{x} - \sqrt{y} \left.\right) \left(\right. \sqrt{x} + \sqrt{y} \left.\right)} . \frac{2 \sqrt{x}}{\sqrt{x} - \sqrt{y}}\)

\(= \frac{- \left(\right. \sqrt{x} - \sqrt{y} \left.\right)^{2}}{2 \left(\right. \sqrt{x} - \sqrt{y} \left.\right) \left(\right. \sqrt{x} + \sqrt{y} \left.\right)} . \frac{2 \sqrt{x}}{\sqrt{x} - \sqrt{y}}\)

\(= \frac{- \sqrt{x}}{\sqrt{x} + \sqrt{y}}\)

b) Ta có \(\frac{x}{y} = \frac{4}{9}\) suy ra \(y = \frac{9}{4} x\)

Do đó \(P = \frac{- \sqrt{x}}{\sqrt{x} + \sqrt{\frac{9 x}{4}}}\)

\(= \frac{- \sqrt{x}}{\sqrt{x} + \frac{3}{2} \sqrt{x}}\)

\(= \frac{- \sqrt{x}}{\frac{5}{2} \sqrt{x}} = - \frac{2}{5}\).

a) Điều kiện \(x \geq 0\), \(y \geq 0\), \(x \neq y\).

Khi đó ta có:

\(P = \left(\right. \frac{2 \sqrt{x y}}{x - y} - \frac{\sqrt{x} + \sqrt{y}}{2 \sqrt{x} - 2 \sqrt{y}} \left.\right) . \frac{2 \sqrt{x}}{\sqrt{x} - \sqrt{y}}\)

\(= \frac{4 \sqrt{x y} - \left(\right. \sqrt{x} + \sqrt{y} \left.\right)^{2}}{2 \left(\right. \sqrt{x} - \sqrt{y} \left.\right) \left(\right. \sqrt{x} + \sqrt{y} \left.\right)} . \frac{2 \sqrt{x}}{\sqrt{x} - \sqrt{y}}\)

\(= \frac{4 \sqrt{x y} - x - 2 \sqrt{x y} - y}{2 \left(\right. \sqrt{x} - \sqrt{y} \left.\right) \left(\right. \sqrt{x} + \sqrt{y} \left.\right)} . \frac{2 \sqrt{x}}{\sqrt{x} - \sqrt{y}}\)

\(= \frac{- \left(\right. \sqrt{x} - \sqrt{y} \left.\right)^{2}}{2 \left(\right. \sqrt{x} - \sqrt{y} \left.\right) \left(\right. \sqrt{x} + \sqrt{y} \left.\right)} . \frac{2 \sqrt{x}}{\sqrt{x} - \sqrt{y}}\)

\(= \frac{- \sqrt{x}}{\sqrt{x} + \sqrt{y}}\)

b) Ta có \(\frac{x}{y} = \frac{4}{9}\) suy ra \(y = \frac{9}{4} x\)

Do đó \(P = \frac{- \sqrt{x}}{\sqrt{x} + \sqrt{\frac{9 x}{4}}}\)

\(= \frac{- \sqrt{x}}{\sqrt{x} + \frac{3}{2} \sqrt{x}}\)

\(= \frac{- \sqrt{x}}{\frac{5}{2} \sqrt{x}} = - \frac{2}{5}\).

a) Điều kiện \(x \geq 0\), \(y \geq 0\), \(x \neq y\).

Khi đó ta có:

\(P = \left(\right. \frac{2 \sqrt{x y}}{x - y} - \frac{\sqrt{x} + \sqrt{y}}{2 \sqrt{x} - 2 \sqrt{y}} \left.\right) . \frac{2 \sqrt{x}}{\sqrt{x} - \sqrt{y}}\)

\(= \frac{4 \sqrt{x y} - \left(\right. \sqrt{x} + \sqrt{y} \left.\right)^{2}}{2 \left(\right. \sqrt{x} - \sqrt{y} \left.\right) \left(\right. \sqrt{x} + \sqrt{y} \left.\right)} . \frac{2 \sqrt{x}}{\sqrt{x} - \sqrt{y}}\)

\(= \frac{4 \sqrt{x y} - x - 2 \sqrt{x y} - y}{2 \left(\right. \sqrt{x} - \sqrt{y} \left.\right) \left(\right. \sqrt{x} + \sqrt{y} \left.\right)} . \frac{2 \sqrt{x}}{\sqrt{x} - \sqrt{y}}\)

\(= \frac{- \left(\right. \sqrt{x} - \sqrt{y} \left.\right)^{2}}{2 \left(\right. \sqrt{x} - \sqrt{y} \left.\right) \left(\right. \sqrt{x} + \sqrt{y} \left.\right)} . \frac{2 \sqrt{x}}{\sqrt{x} - \sqrt{y}}\)

\(= \frac{- \sqrt{x}}{\sqrt{x} + \sqrt{y}}\)

b) Ta có \(\frac{x}{y} = \frac{4}{9}\) suy ra \(y = \frac{9}{4} x\)

Do đó \(P = \frac{- \sqrt{x}}{\sqrt{x} + \sqrt{\frac{9 x}{4}}}\)

\(= \frac{- \sqrt{x}}{\sqrt{x} + \frac{3}{2} \sqrt{x}}\)

\(= \frac{- \sqrt{x}}{\frac{5}{2} \sqrt{x}} = - \frac{2}{5}\).