50đ nếu giải được ạ có trước 6h sáng mai là được ạ=(((((

Đáp án:

a. Khi x = 16 (TMĐKXĐ)

=> A = $\frac{\sqrt{16} -2}{\sqrt{16}+3}$

<=> A = $\frac{4-2}{4+3}$

<=> A = $\frac{2}{7}$ 

b. B = $\frac{\sqrt{x}-1}{\sqrt{x}-3}$ - $\frac{5\sqrt{x}-3}{x-9}$

<=> B = $\frac{(\sqrt{x}-1)(\sqrt{x}+3)}{(\sqrt{x}-3)(\sqrt{x}+3)}$ - $\frac{5\sqrt{x}-3}{(\sqrt{x}-3)(\sqrt{x}+3)}$

<=> B = $\frac{x+3\sqrt{x}-\sqrt{x}-3}{(\sqrt{x}-3)(\sqrt{x}+3)}$ - $\frac{5\sqrt{x}-3}{(\sqrt{x}-3)(\sqrt{x}+3)}$

<=> B = $\frac{x+2\sqrt{x}+3}{(\sqrt{x}-3)(\sqrt{x}-3)}$ - $\frac{5\sqrt{x}-3}{(\sqrt{x}-3)(\sqrt{x}+3)}$

<=> B = $\frac{x+2\sqrt{x}-3-(5\sqrt{x}-3)}{(\sqrt{x}-3)(\sqrt{x}+3)}$

<=> B = $\frac{x+2\sqrt{x}-3-5\sqrt{x}+3}{(\sqrt{x}-3)(\sqrt{x}+3)}$

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

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

<=> $\frac{A}{B}$ = A : B = $\frac{\sqrt{x} -2}{\sqrt{x}+3}$ : $\frac{\sqrt{x}(\sqrt{x}-3)}{(\sqrt{x}-3)(\sqrt{x}+3)}$

<=> $\frac{A}{B}$ = $\frac{\sqrt{x} -2}{\sqrt{x}+3}$ . $\frac{(\sqrt{x}-3)(\sqrt{x}+3)}{\sqrt{x}(\sqrt{x}-3)}$

<=> $\frac{A}{B}$ = $\frac{\sqrt{x} -2}{\sqrt{x}+3}$ . $\frac{(\sqrt{x}-3)(\sqrt{x}+3)}{\sqrt{x}(\sqrt{x}-3)}$

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

<=> $\frac{A}{B}$ = $\frac{\sqrt{x} -2}{\sqrt{x}}$ 

Vậy $\frac{A}{B}$ = $\frac{\sqrt{x} -2}{\sqrt{x}}$