Question

[zZz Nguyễn zZz]

Cho \(A=\left(\frac{x-\sqrt{x}+7}{x-4}+\frac{1}{\sqrt{x}-2}\right):\left(\frac{\sqrt{x}+2}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+2}-\frac{2\sqrt{x}}{x-4}\right)\)

a) ĐKXĐ , Rút Gọn

b)So sánh A với 1/A

Answer 1

[Lê Anh Duy]

Cái này là toán lớp 9 chứ.

a)
ĐKXĐ : \(x\ne\pm4\)

\(A=\left(\frac{x-\sqrt{x}+7}{x-4}+\frac{\sqrt{x}+2}{x-4}\right):\left(\frac{\left(\sqrt{x}+2\right)^2}{x-4}-\frac{\left(\sqrt{x}-2\right)^2}{x-4}-\frac{2\sqrt{x}}{x-4}\right)\)

\(=\left(\frac{x-\sqrt{x}+7+\sqrt{x}+2}{x-4}\right):\left(\frac{x+4\sqrt{x}+4-x+4\sqrt{x}-4-2\sqrt{x}}{x-4}\right)\)

\(=\frac{x+9}{x-4}\cdot\frac{x-4}{6\sqrt{x}}=\frac{x+9}{6\sqrt{x}}\)

b)

Ta có

\(x+9-6\sqrt{x}=\left(\sqrt{x}-3\right)^2\ge0\)
\(\Rightarrow x+9\ge6\sqrt{x}\)

\(\Rightarrow\frac{x+9}{6\sqrt{x}}\ge1\)

\(\Leftrightarrow A\ge1\)

\(\Leftrightarrow\frac{1}{A}\le1\)

\(\Rightarrow A\ge\frac{1}{A}\)