Question

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

a) Rút gọn

b) Tìm giá trị thực của H để x = \(14-6\sqrt{5}\)

2. Cho P = \(\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}+\frac{x+1}{\sqrt{x}}\)

a) Rút gọn

Answer 1

Bài 1:

ĐKXĐ: \(x\geq 0; x\neq 9\)

a)

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

\(=\frac{x\sqrt{x}-3-2(x-6\sqrt{x}+9)-(x+4\sqrt{x}+3)}{(\sqrt{x}+1)(\sqrt{x}-3)}\)

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

b)

\(x=14-6\sqrt{5}=5+3^2-2.3\sqrt{5}=(3-\sqrt{5})^2\Rightarrow \sqrt{x}=3-\sqrt{5}\)

\(\Rightarrow H=\frac{x+8}{\sqrt{x}+1}=\frac{14-6\sqrt{5}+8}{3-\sqrt{5}+1}=\frac{22-6\sqrt{5}}{4-\sqrt{5}}\)

Answer 2

Bài 2:

ĐKXĐ: \(x>0; x\neq 1\)

\(P=\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}+\frac{x+1}{\sqrt{x}}\)

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

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

Answer 3

Vậy cho mik hỏi làm sao để tìm Min của H trong bài 1 vậy ?

Answer 4

Mong mọi người chiếu cố cho !