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

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

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

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

\(= \frac{\sqrt{x} + 1}{\sqrt{x}}\).

b) \(2 P = 2 \sqrt{x} + 5\)

\(\frac{2 \left(\right. \sqrt{x} + 1 \left.\right)}{\sqrt{x}} = 2 \sqrt{x} + 5\)

\(2 \left(\right. \sqrt{x} + 1 \left.\right) = 2 x + 5 \sqrt{x}\)

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

\(\left(\right. 2 \sqrt{x} - 1 \left.\right) \left(\right. \sqrt{x} + 2 \left.\right) = 0\)

\(2 \sqrt{x} - 1 = 0\)

\(x = \frac{1}{4}\) (thỏa mãn).

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

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

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

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

\(= \frac{\sqrt{x} + 1}{\sqrt{x}}\).

b) \(2 P = 2 \sqrt{x} + 5\)

\(\frac{2 \left(\right. \sqrt{x} + 1 \left.\right)}{\sqrt{x}} = 2 \sqrt{x} + 5\)

\(2 \left(\right. \sqrt{x} + 1 \left.\right) = 2 x + 5 \sqrt{x}\)

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

\(\left(\right. 2 \sqrt{x} - 1 \left.\right) \left(\right. \sqrt{x} + 2 \left.\right) = 0\)

\(2 \sqrt{x} - 1 = 0\)

\(x = \frac{1}{4}\) (thỏa mãn).

a) Với \(x = 9\) (thỏa mãn điều kiện) thì \(\sqrt{x} = 3\).

Thay vào biểu thức \(P\) ta có: \(P = \frac{9 + 3}{3 - 2} = 12\).

b) \(Q = \frac{\sqrt{x} - 1}{\sqrt{x} + 2} + \frac{5 \sqrt{x} - 2}{x - 4}\)

\(= \frac{\left(\right. \sqrt{x} - 1 \left.\right) \left(\right. \sqrt{x} - 2 \left.\right)}{\left(\right. \sqrt{x} + 2 \left.\right) \left(\right. \sqrt{x} - 2 \left.\right)} + \frac{5 \sqrt{x} - 2}{\left(\right. \sqrt{x} + 2 \left.\right) \left(\right. \sqrt{x} - 2 \left.\right)}\)

\(= \frac{x - 3 \sqrt{x} + 2 + 5 \sqrt{x} \&\text{nbsp}; - 2}{\left(\right. \sqrt{x} + 2 \left.\right) \left(\right. \sqrt{x} - 2 \left.\right)}\)

\(= \frac{x \&\text{nbsp}; + 2 \sqrt{x}}{\left(\right. \sqrt{x} + 2 \left.\right) \left(\right. \sqrt{x} - 2 \left.\right)}\)

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

c) Ta có \(\frac{P}{Q} = \frac{x + 3}{\sqrt{x} - 2} : \frac{\sqrt{x}}{\sqrt{x} - 2}\)

\(= \frac{x + 3}{\sqrt{x}} = \sqrt{x} + \frac{3}{\sqrt{x}}\).

Áp dụng bất đẳng thức Cauchy ta có: \(\sqrt{x} + \frac{3}{\sqrt{x}} \geq 2. \sqrt{\frac{\sqrt{x} . 3}{\sqrt{x}}} = 2 \sqrt{3}\).

Vậy giá trị nhỏ nhất của \(\frac{P}{Q} = 2 \sqrt{3}\), đẳng thức xảy ra khi \(\sqrt{x} = \frac{3}{\sqrt{x}}\) hay \(x = 3\).

a) Điều kiện: \(x \neq 4 ,\) \(x \geq 0\).

Ta có \(x = 9\) (thỏa mãn điều kiện) thì \(\sqrt{x} = 3\)

Thay vào biểu thức \(A\) ta được: \(A = \frac{3}{3 - 2} = 3\).

b) \(T = A - B = \frac{\sqrt{x}}{\sqrt{x} - 2} - \frac{2}{\sqrt{x} + 2} - \frac{4 \sqrt{x}}{x - 4}\)

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

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

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

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

c) \(\frac{\sqrt{x} - 2}{\sqrt{x} + 2}\)

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

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

\(= 1 - \frac{4}{\sqrt{x} + 2}\)

Vậy để \(T\) nguyên thì \(\frac{4}{\sqrt{x} + 2} \in \mathbb{Z}\)

hay \(\sqrt{x} + 2\) là ước của \(4\).

Vậy \(x = 0\) là giá trị cần tìm.

a) \(P = \frac{3}{\sqrt{x} + 1} - \frac{1}{\sqrt{x} - 1} - \frac{\sqrt{x} - 5}{x - 1}\)

\(= \frac{3 \left(\right. \sqrt{x} - 1 \left.\right) - \left(\right. \sqrt{x} + 1 \left.\right) - \left(\right. \sqrt{x} - 5 \left.\right)}{\left(\right. \sqrt{x} + 1 \left.\right) \left(\right. \sqrt{x} - 1 \left.\right)}\)

\(= \frac{3 \sqrt{x} - 3 - \sqrt{x} - 1 - \sqrt{x} + 5}{\left(\right. \sqrt{x} - 1 \left.\right) \left(\right. \sqrt{x} + 1 \left.\right)}\)

\(= \frac{\sqrt{x} + 1}{\left(\right. \sqrt{x} - 1 \left.\right) \left(\right. \sqrt{x} + 1 \left.\right)}\)

\(= \frac{1}{\sqrt{x} - 1}\).

b) Chú ý \(\sqrt{x} = \sqrt{\left(\right. 4 - 2 \sqrt{2} \left.\right)^{2}} = 4 - 2 \sqrt{2}\).

Vậy \(P = \frac{1}{4 - 2 \sqrt{2} - 1}\)

\(= \frac{1}{3 - 2 \sqrt{2}}\)

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

\(= 3 + 2 \sqrt{2}\).

a) Ta có:

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

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

\(= 2 \sqrt{x} + 1\).

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

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

\(= x - 1\).

b) Để \(P = Q\) thì \(2 \sqrt{x} + 1 = x - 1\)

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

Coi phương trình là phương trình bậc hai của \(\sqrt{x}\), chú ý chọn nghiệm dương của phương trình ta được \(\sqrt{x} = 1 + \sqrt{3}\) nên \(x = 4 + 2 \sqrt{3}\) (thỏa mãn).

Suy ra \(x = 4 + 2 \sqrt{3}\) thì \(P = Q\).

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.

A=3+23⋅3​=59​

2: \(B = \frac{x + 4}{x - 4} - \frac{2}{\sqrt{x} - 2}\)

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

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

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

3: \(A - B < \frac{3}{2}\)

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

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

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

=>\(\frac{\sqrt[]{x} - 6}{2 \left(\right. \sqrt{x} + 2 \left.\right)} < 0\)

=>\(\sqrt{x} - 6 < 0\)

=>\(\sqrt{x} < 6\)

=>0<=x<36

mà x là số nguyên dương lớn nhất thỏa mãn

nên x=35