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

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

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

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

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

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) Điều kiện \(x > 0\).

Khi đó ta có:

\(P = \left(\right. \frac{1}{\sqrt{x} + 3} + \frac{3}{x \sqrt{x} - 9 \sqrt{x}} \left.\right) : \left(\right. \frac{\sqrt{x}}{\sqrt{x} + 3} - \frac{3 \sqrt{x} - 3}{x + 3 \sqrt{x}} \left.\right)\)

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

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

b) Ta có \(P > 1\)

\(\frac{1}{\sqrt{x} - 3} > 1\)

\(\frac{1}{\sqrt{x} - 3} - 1 > 0\)

\(\frac{1 - \sqrt{x} + 3}{\sqrt{x} - 3} > 0\)

\(\frac{\sqrt{x} - 4}{\sqrt{x} - 3} < 0\)

Suy ra \(\sqrt{x} - 4 > 0\) và \(\sqrt{x} - 3 < 0\) hoặc \(\sqrt{x} - 4 < 0\) và \(\sqrt{x} - 3 > 0\)

\(9 < x < 16\) (thỏa mãn điều kiện).

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) 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 \(x = 35\).

P=x+2x​x+2​−x​1​+x​+21​

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

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

P=(x​+1x​​+x​−1x​​)(x​−x​1​)

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

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

B=(x​+1x​​−x​x​−1​):x​+1x​​

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

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