a: Thay x=36 vào A, ta được:

\(A=\frac{\sqrt{36}+4}{\sqrt{36}+2}=\frac{6+4}{6+2}=\frac{10}{8}=\frac54\)

b: \(B=\left(\frac{\sqrt{x}}{\sqrt{x}+4}+\frac{4}{\sqrt{x}-4}\right):\frac{x+16}{\sqrt{x}+2}\)

\(=\frac{\sqrt{x}\left(\sqrt{x}-4\right)+4\left(\sqrt{x}+4\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}\cdot\frac{\sqrt{x}+2}{x+16}\)

\(=\frac{x-4\sqrt{x}+4\sqrt{x}+16}{x-16}\cdot\frac{\sqrt{x}+2}{x+16}=\frac{\sqrt{x}+2}{x-16}\)

c: Đặt P=B(A-1)

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

\(=\frac{\sqrt{x}+2}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}\cdot\frac{2}{\sqrt{x}+2}=\frac{2}{x-16}\)

Để P là số nguyên thì 2⋮x-16

=>x-16∈{1;-1;2;-2}

=>x∈{17;15;18;14}

a) \(A=\dfrac{\sqrt{x}+4}{\sqrt{x}+2}=\dfrac{\sqrt{x}+2+2}{\sqrt{x}+2}=1+\dfrac{2}{\sqrt{x}+2}=1+\dfrac{2\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=1+\dfrac{2\left(\sqrt{x}-2\right)}{x-4}\)Thay x = 6 vào A ta có :

\(A=1+\dfrac{2\left(\sqrt{6}-2\right)}{6-4}=1+\sqrt{6}-2=\sqrt{6}-1\)

b)

\(B=\left(\dfrac{\sqrt{x}}{\sqrt{x}+4}+\dfrac{4}{\sqrt{x}-4}\right):\dfrac{x+16}{\sqrt{x}+2}=\dfrac{\sqrt{x}\left(\sqrt{x}-4\right)+4\left(\sqrt{x}+4\right)}{x-16}\cdot\dfrac{\sqrt{x}+2}{x+16}=\dfrac{x-4\sqrt{x}+4\sqrt{x}+16}{x-16}\cdot\dfrac{\sqrt{x}+2}{x+16}=\dfrac{x+16}{x-16}\cdot\dfrac{\sqrt{x}+2}{x+16}=\dfrac{\sqrt{x}+2}{x-16}\)

a/ \(P=12\)

b/ \(Q=\frac{\sqrt{x}}{\sqrt{x}-2}\)
c/ Ta có:

\(\frac{P}{Q}=\frac{\frac{x+3}{\sqrt{x}-2}}{\frac{\sqrt{x}}{\sqrt{x}-2}}=\frac{x+3}{\sqrt{x}}\ge\frac{2\sqrt{3x}}{\sqrt{x}}=2\sqrt{3}\)
Dấu = xảy ra khi x = 3 (thỏa tất cả các điều kiện )

a. Thay x = 3 vào biểu thức P ta được :

\(p=\frac{x+3}{\sqrt{x}-2}=\frac{9+3}{\sqrt{9}-2}=12\)

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

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

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

\(=\frac{x-3\sqrt{x}+2+5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

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

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

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

c, Ta có :

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

Vậy GTNN \(\frac{P}{Q}=2\sqrt{3}\) khi và chỉ khi \(x=3\)

a: \(B=\frac{\sqrt{x}}{x+\sqrt{x}}\)

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

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

P=A:B

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

Thay x=4 vào P, ta được:

\(P=\frac{4+2+1}{2}=\frac72\)

b: A<=3B

=>\(\frac{x+\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\le\frac{3}{\sqrt{x}+1}=\frac{3\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

=>\(x+\sqrt{x}+1\le3\sqrt{x}\)

=>\(x-2\sqrt{x}+1\le0\)

=>\(\left(\sqrt{x}-1\right)^2\le0\)

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

=>x=1(nhận)c

c: \(B-1=\frac{1}{\sqrt{x}+1}-1=\frac{1-\sqrt{x}-1}{\sqrt{x}+1}=\frac{-\sqrt{x}}{\sqrt{x}+1}<0\forall x\) thỏa mãn ĐKXĐ
=>B<1∀x thỏa mãn ĐKXĐ