Question

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

a) Rút gọn A

b) Tính A với x=14-6\(\sqrt{5}\)

c) Tìm Min A

Answer 1

Lời giải:
ĐKXĐ: $x\geq 0; x\neq 9$

a. \(A=\frac{x\sqrt{x}-3}{(\sqrt{x}+1)(\sqrt{x}-3)}-\frac{2(\sqrt{x}-3)^2}{(\sqrt{x}+1)(\sqrt{x}-3)}-\frac{(\sqrt{x}+3)(\sqrt{x}+1)}{(\sqrt{x}+1)(\sqrt{x}-3)}\)

\(=\frac{x\sqrt{x}-3x+8\sqrt{x}-24}{(\sqrt{x}+1)(\sqrt{x}-3)}=\frac{(\sqrt{x}-3)(x+8)}{(\sqrt{x}+1)(\sqrt{x}-3)}=\frac{x+8}{\sqrt{x}+1}\)

b.

\(14-6\sqrt{5}=(3-\sqrt{5})^2\Rightarrow \sqrt{x}=3-\sqrt{5}\)

\(A=\frac{14-6\sqrt{5}+8}{3-\sqrt{5}+1}=\frac{22-6\sqrt{5}}{4-\sqrt{5}}=\frac{58-2\sqrt{5}}{11}\)

c. 

Áp dụng BĐT Cô-si:
$x+4\geq 4\sqrt{x}\Rightarrow x+8\geq 4(\sqrt{x}+1)$

$\Rightarrow A=\frac{x+8}{\sqrt{x}+1}\geq 4$

Vậy $A_{\min}=4$. Giá trị này đạt tại $x=4$