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) \(A = \left(\right. \frac{4 \sqrt{x}}{2 + \sqrt{x}} + \frac{8 x}{4 - x} \left.\right) : \left(\right. \frac{\sqrt{x} - 1}{x - 2 \sqrt{x}} - \frac{2}{\sqrt{x}} \left.\right)\)

\(= \left[\right. \frac{4 \sqrt{x} \left(\right. 2 - \sqrt{x} \left.\right) + 8 x}{\left(\right. 2 + \sqrt{x} \left.\right) \left(\right. 2 - \sqrt{x} \left.\right)} \left]\right. : \left[\right. \frac{\sqrt{x} - 1 - 2 \left(\right. \sqrt{x} - 2 \left.\right)}{\sqrt{x} \left(\right. \sqrt{x} - 2 \left.\right)} \left]\right.\)

\(= \frac{8 \sqrt{x} + 4 x}{\left(\right. 2 + \sqrt{x} \left.\right) \left(\right. 2 - \sqrt{x} \left.\right)} . \frac{\sqrt{x} \left(\right. \sqrt{x} - 2 \left.\right)}{3 - \sqrt{x}}\)

\(= \frac{4 \sqrt{x} \left(\right. 2 + \sqrt{x} \left.\right)}{\left(\right. 2 + \sqrt{x} \left.\right) \left(\right. 2 - \sqrt{x} \left.\right)} . \frac{- \sqrt{x} \left(\right. 2 - \sqrt{x} \left.\right)}{3 - \sqrt{x}}\)

\(= \frac{4 x}{\sqrt{x} - 3}\).

b) Để \(A = - 2\) thì \(\frac{4 x}{\sqrt{x} - 3} = - 2\)

\(\frac{4 x}{\sqrt{x} - 3} + 2 = 0\)

\(\frac{4 x + 2 \left(\right. \sqrt{x} - 3 \left.\right)}{\sqrt{x} - 3} = 0\)

\(4 x + 2 \sqrt{x} - 6 = 0\)

\(\sqrt{x} = 1\) hoặc \(\sqrt{x} = - \frac{3}{2}\) (vô lí)

Suy ra \(x = 1\) (thỏa mãn).

a) Điều kiện xác định: \(a > 0\); \(a \neq 1\) và \(a \neq 2\).

\(P = \left(\right. \frac{1}{\sqrt{a} - 1} - \frac{1}{\sqrt{a}} \left.\right) : \left(\right. \frac{\sqrt{a} + 1}{\sqrt{a} - 2} - \frac{\sqrt{a} + 2}{\sqrt{a} - 1} \left.\right)\)

\(= \frac{\sqrt{a} - \sqrt{a} + 1}{\left(\right. \sqrt{a} - 1 \left.\right) \sqrt{a}} : \left[\right. \frac{\left(\right. \sqrt{a} + 1 \left.\right) \left(\right. \sqrt{a} - 1 \left.\right)}{\left(\right. \sqrt{a} - 1 \left.\right) \left(\right. \sqrt{a} - 2 \left.\right)} - \frac{\left(\right. \sqrt{a} + 2 \left.\right) \left(\right. \sqrt{a} - 2 \left.\right)}{\left(\right. \sqrt{a} - 1 \left.\right) \left(\right. \sqrt{a} - 2 \left.\right)} \left]\right.\)

\(= \frac{1}{\left(\right. \sqrt{a} - 1 \left.\right) \sqrt{a}} : \frac{a - 1 - a + 4}{\left(\right. \sqrt{a} - 1 \left.\right) \left(\right. \sqrt{a} - 2 \left.\right)}\)

\(= \frac{1}{\left(\right. \sqrt{a} - 1 \left.\right) \sqrt{a}} . \frac{\left(\right. \sqrt{a} - 1 \left.\right) \left(\right. \sqrt{a} - 2 \left.\right)}{3}\)

\(= \frac{\sqrt{a} - 2}{3 \sqrt{a}}\).

b) Để \(P > \frac{1}{6}\) thì \(\frac{\sqrt{a} - 2}{3 \sqrt{a}} > \frac{1}{6}\).

Vì \(\sqrt{a} > 0\) thỏa mãn điều kiện xác định nên để \(\frac{\sqrt{a} - 4}{6 \sqrt{a}} > 0\) thì \(\sqrt{a} - 4 > 0.\)

\(\sqrt{a} > 4\)

\(a > 16\).

Kết hợp điều kiện, ta có \(a > 16\) là giá trị cần tìm.

1) Thay \(x = 9\) (TMĐK) vào biểu thức \(A\), ta có:

\(A = \frac{3 \sqrt{9}}{\sqrt{9} + 2} = \frac{9}{5}\).

2) Ta có: \(B = \frac{x + 4}{\left(\right. \sqrt{x} + 2 \left.\right) \left(\right. \sqrt{x} - 2 \left.\right)} - \frac{2}{\sqrt{x} - 2}\)

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

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

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

\(= \frac{\sqrt{x}}{\sqrt{x} + 2}\).

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

\(A - B < \frac{3}{2}\) khi \(\frac{2 \sqrt{x}}{\sqrt{x} + 2} < \frac{3}{2}\)

\(4 \sqrt{x} < 3 \sqrt{x} + 6\) vì \(x \geq 0\) nên \(\sqrt{x} + 2 > 0\)

\(\sqrt{x} < 6\)

\(x < 36\)

Kết hợp với điều kiện và yêu cầu của bài toán, suy ra \(x = 35\).

Vậy số nguyên dương \(x\) lớn nhất thỏa mãn \(A - B < \frac{3}{2}\) là \(x = 35\).