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

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

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

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

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

b: \(\frac{x}{y} = \frac{4}{9}\)

=>\(\frac{x}{4} = \frac{y}{9} = k\)

=>x=4k; y=9k

\(P = \frac{- \sqrt{x}}{\sqrt{x} + \sqrt{y}} = \frac{- \sqrt{4 k}}{\sqrt{4 k} + \sqrt{9 k}} = \frac{- 2 \sqrt{k}}{2 \sqrt{k} + 3 \sqrt{k}} = - \frac{2}{5}\)

ĐKXĐ: x>0; x<>9

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

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

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

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

b: P>1

=>P-1>0

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

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

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

=>\(3 < \sqrt{x} < 4\)

=>9<x<16

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

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

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

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

b: \(2 P = 2 \sqrt{x} + 5\)

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

=>\(2 x + 5 \sqrt{x} - 2 \sqrt{x} - 2 = 0\)

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

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

mà \(\sqrt{x} + 2 > = 2 > 0 \forall x\) thỏa mãn ĐKXĐ

nên \(2 \sqrt{x} - 1 = 0\)

=>\(\sqrt{x} = \frac{1}{2}\)

=>\(x = \frac{1}{4} \left(\right. n h ậ n \left.\right)\)

a: Thay x=9 vào P, ta được:

\(P = \frac{9 + 3}{\sqrt{9} - 2} = \frac{12}{3 - 2} = \frac{12}{1} = 12\)

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

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

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

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

c: Đặt A=P:Q

\(= \frac{x + 3}{\sqrt{x} - 2} : \frac{\sqrt{x}}{\sqrt{x} - 2} = \frac{x + 3}{\sqrt{x}} = \sqrt{x} + \frac{3}{\sqrt{x}} > = 2 \cdot \sqrt{\sqrt{x} \cdot \frac{3}{\sqrt{x}}} = 2 \sqrt{3}\) với mọi x thỏa mãn ĐKXĐ

Dấu '=' xảy ra khi \(\left(\left(\right. \sqrt{x} \left.\right)\right)^{2} = 3\)

=>x=3(nhận)

ĐKXĐ: x>=0; x<>4

a: Thay x=9 vào A, ta được:

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

b: T=A-B

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

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

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

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

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

c: Để T nguyên thì \(\sqrt{x} - 2 \sqrt{x} + 2\)

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

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

mà \(\sqrt{x} + 2 > = 2 \forall x\) thỏa mãn ĐKXĐ

 nên \(\sqrt{x} + 2 \in \left{\right. 2 ; 4 \left.\right}\)

=>\(x \in \left{\right. 0 ; 4 \left.\right}\)

Kết hợp ĐKXĐ, ta được: x=0

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

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

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

\(= \frac{3 \sqrt{x} - 3 - 2 \sqrt{x} + 4}{\left(\right. \sqrt{x} - 1 \left.\right) \left(\right. \sqrt{x} + 1 \left.\right)} = \frac{\sqrt{x} + 1}{\left(\right. \sqrt{x} - 1 \left.\right) \left(\right. \sqrt{x} + 1 \left.\right)} = \frac{1}{\sqrt{x} - 1}\)

b: \(x = 24 - 16 \sqrt{2} = 8 \left(\right. 3 - 2 \sqrt{2} \left.\right) = 8 \left(\left(\right. \sqrt{2} - 1 \left.\right)\right)^{2}\)

Thay \(x = 8 \left(\left(\right. \sqrt{2} - 1 \left.\right)\right)^{2}\) vào P, ta được:

\(P = \frac{1}{\sqrt{8 \left(\left(\right. \sqrt{2} - 1 \left.\right)\right)^{2}} - 1}\)

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

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

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

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

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

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

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

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

b: P=Q

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

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

=>\(x - 2 \sqrt{x} + 1 = 3\)

=>\(\left(\left(\right. \sqrt{x} - 1 \left.\right)\right)^{2} = 3\)

mà \(\sqrt{x} - 1 > = - 1\) với mọi x thỏa mãn ĐKXĐ

nên \(\sqrt{x} - 1 = \sqrt{3}\)

=>\(\sqrt{x} = 1 + \sqrt{3}\)

=>\(x = \left(\left(\right. 1 + \sqrt{3} \left.\right)\right)^{2} = 4 + 2 \sqrt{3} \left(\right. n h ậ n \left.\right)\)

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

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

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

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

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

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

b: A=-2

=>\(4 x = - 2 \left(\right. \sqrt{x} - 3 \left.\right) = - 2 \sqrt{x} + 6\)

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

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

=>\(\left(\right. 2 \sqrt{x} + 3 \left.\right) \left(\right. \sqrt{x} - 1 \left.\right) = 0\)

mà \(2 \sqrt{x} + 3 > = 3 > 0 \forall x\) thỏa mãn ĐKXĐ

nên \(\sqrt{x} - 1 = 0\)

=>x=1(nhận)

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

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

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

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

b: P>1/6

=>P-1/6>0

=>\(\frac{\sqrt{a} - 2}{3 \sqrt{a}} - \frac{1}{6} > 0\)

=>\(\frac{6 \left(\right. \sqrt{a} - 2 \left.\right) - 3 \sqrt{a}}{18 \sqrt{a}} > 0\)

=>\(6 \left(\right. \sqrt{a} - 2 \left.\right) - 3 \sqrt{a} > 0\)

=>\(3 \sqrt{a} - 12 > 0\)

=>\(\sqrt{a} > 4\)
=>a>16

1: Thay x=9 vào A, ta được:

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

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