Question

cho bt \(A=\left(\frac{\sqrt{x}-1}{x-4}-\frac{\sqrt{x}+1}{x+4\sqrt{x}+4}\right):\frac{x\sqrt{x}}{\left(4-x\right)^2}\)

1, rut gọn A

2, tính giá trị của A tại \(x=4+2\sqrt{3}\)

3,tìm x de A\(\ge\frac{1}{4}\)

Answer 1

ĐK: \(x>0,x\ne4\)

1, \(A=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)-\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)^2}.\frac{\left(x-4\right)^2}{\sqrt{x}^3}\)

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

\(A=\frac{2}{\sqrt{x}+2}.\frac{x-4}{x}\)

\(A=\frac{2\sqrt{x}-4}{x}\)

2, \(x=\left(\sqrt{3}+1\right)^2\)

Thay \(x=\left(\sqrt{3}+1\right)^2\):

\(A=\frac{2\sqrt{3}-2}{\left(\sqrt{3}+1\right)^2}\)

3, \(A\ge\frac{1}{4}\Rightarrow\)\(\frac{2\sqrt{x}-4}{x}-\frac{1}{4}\ge0\)

\(\Leftrightarrow\frac{8\sqrt{x}-16-x}{4x}\ge0\)

​\(\Rightarrow x-8\sqrt{x}+16\le0\)

\(\Leftrightarrow\sqrt{x}=4\Leftrightarrow x=16\left(TM\right)\)

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