c,M =  \(\dfrac{A}{B}\) = \(\dfrac{\sqrt{x}-4}{\sqrt{x}+5}\) :  \(\dfrac{\sqrt{x}+3}{\sqrt{x}+5}\) 

   M =  \(\dfrac{A}{B}\) = \(\dfrac{\sqrt{x}-4}{\sqrt{x}+5}\) \(\times\) \(\dfrac{\sqrt{x}+5}{\sqrt{x}+3}\) 

   M =  \(\dfrac{A}{B}\) = \(\dfrac{\sqrt{x}-4}{\sqrt{x}+3}\) = \(\dfrac{\sqrt{x}+3-7}{\sqrt{x}+3}\)

 M = 1  - \(\dfrac{7}{\sqrt{x}+3}\) 

 M \(\in\) Z ⇔ 7 ⋮ \(\sqrt{x}\) + 3 vì \(\sqrt{x}\) ≥ 0 ⇒ \(\sqrt{x}\) + 3 ≥ 3 ⇒ 0< \(\dfrac{7}{\sqrt{x}+3}\) ≤ \(\dfrac{7}{3}\)

⇒ M Đạt giá trị nguyên lớn nhất ⇔ \(\dfrac{7}{\sqrt{x}+3}\) đạt giá trị nguyên nhỏ nhất ⇔ \(\dfrac{7}{\sqrt{x}+3}\) = 1 ⇔ \(\sqrt{x}\) + 3  = 7 ⇔ \(\sqrt{x}\) = 4 ⇔ \(x\) = 16 

Mnguyên(max)  = 1 - 1 = 0 xảy ra khi \(x\) = 16

b) Thay x=49 vào A, ta được:

\(A=\dfrac{7-1}{7-5}=\dfrac{6}{2}=3\)

a) Ta có: \(B=\dfrac{\sqrt{x}+3}{\sqrt{x}+1}+\dfrac{5}{\sqrt{x}-1}+\dfrac{4}{x-1}\)

\(=\dfrac{x+2\sqrt{x}-3+5\sqrt{x}+5+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{x+7\sqrt{x}+6}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{\sqrt{x}+6}{\sqrt{x}-1}\)

\(A=\dfrac{\sqrt{x}}{\sqrt{x}-5}-\dfrac{10\sqrt{x}}{x-25}-\dfrac{5}{\sqrt{x}+5}\left(x\ge0;x\ne25\right)\)

Để \(A=\dfrac{2\sqrt{x}}{3}\) thì:

\(\dfrac{\sqrt{x}-5}{\sqrt{x}+5}=\dfrac{2\sqrt{x}}{3}\)

\(\Leftrightarrow3\sqrt{x}-15=2x+10\sqrt{x}\)

\(\Leftrightarrow2x+10\sqrt{x}-3\sqrt{x}+15=0\)

\(\Leftrightarrow2x+7\sqrt{x}+15=0\) 

Mà \(2x+7\sqrt{x}+15>0\) (vì \(x\ge0\))

nên không tìm được giá trị nào của \(x\) thoả mãn \(A=\dfrac{2\sqrt{x}}{3}\)

#\(Toru\)

Có \(A=\dfrac{\sqrt{x}-5}{\sqrt{x}+5}=1-\dfrac{10}{\sqrt{x}+5}\)

Dễ thấy \(\dfrac{10}{\sqrt{x}+5}>0\forall x\Rightarrow A=1-\dfrac{10}{\sqrt{x}+5}< 1\)

=> A < 2

2: \(A=\dfrac{\sqrt{x}-5}{\sqrt{x}+5}=\dfrac{\sqrt{x}+5-10}{\sqrt{x}+5}\)

\(=1-\dfrac{10}{\sqrt{x}+5}\)

\(\sqrt{x}+5>=5\forall x\)

=>\(\dfrac{10}{\sqrt{x}+5}< =\dfrac{10}{5}=2\forall x\)

=>\(-\dfrac{10}{\sqrt{x}+5}>=-2\forall x\)

=>\(-\dfrac{10}{\sqrt{x}+5}+1>=-2+1=-1\forall x\)

Dấu '=' xảy ra khi x=0

Vậy: \(A_{min}=-1\) khi x=0

a: Thay x=9 vào A, ta được:

\(A=\dfrac{3+2}{3-5}=\dfrac{5}{-2}=\dfrac{-5}{2}\)

\(B=\dfrac{3\sqrt{x}-15+20-2\sqrt{x}}{x-25}=\dfrac{\sqrt{x}+5}{x-25}=\dfrac{1}{\sqrt{x}-5}\)

b: Để \(A=B\cdot\left|x-4\right|\) thì \(\left|x-4\right|=\dfrac{A}{B}=\dfrac{\sqrt{x}+2}{\sqrt{x}-5}:\dfrac{1}{\sqrt{x}-5}=\sqrt{x}+2\)

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

\(\Leftrightarrow x-\sqrt{x}-6=0\)

=>x=9

`A)đk:x>=0,x ne 25`

`A=9=>A=(3+2)/(3-5)=-5/2`

`B)B=(3sqrtx-15+20-2sqrtx)/(x-25)`

`=(sqrtx+5)/(x-25)`

`=1/(sqrtx-5)`

`A=B.|x-4|`

`<=>A/B=|x-4|`

`<=>\sqrtx+2=|x-4|`

`<=>\sqrtx+2=(sqrtx+2)|sqrtx-2|`

`<=>|sqrtx-2|=1`

`+)sqrtx-2=1<=>x=9(tm)`

`+)sqrtx-2=-1<=>x=1(tm)`

Vậy `S={1,9}`

a, Thay x=9 vào biểu thức A ta có

\(A=\dfrac{\sqrt{9}+2}{\sqrt{9}-5}\)

\(A=\dfrac{3+2}{3-5}=\dfrac{5}{-2}=-2,5\)

Vậy A =-2,5 khi x=9

a. A=\(\dfrac{\sqrt{x}+2}{\sqrt{x}-5}\)

=\(\dfrac{\sqrt{9}+2}{\sqrt{9}-5}=\dfrac{-5}{2}\)