Question

Mn giúp e với 
E cảm ơn ạ

Answer 1

1, Với x > 0 ; x khác 4 

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

\(=\left(\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)-\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(x-4\right)\left(\sqrt{x}+2\right)}\right):\dfrac{x\sqrt{x}}{\left(x-4\right)^2}\)

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

\(=\dfrac{2\sqrt{x}\left(x-4\right)^2}{x\sqrt{x}\left(x-4\right)\left(\sqrt{x}+2\right)}=\dfrac{2\left(\sqrt{x}-2\right)}{x}\)

2, Ta có \(x=4+2\sqrt{3}\Rightarrow\sqrt{x}=\sqrt{4+2\sqrt{3}}=\sqrt{3}+1\)

Thay vào ta được \(A=\dfrac{2\left(\sqrt{3}-1\right)}{4+2\sqrt{3}}=-5+3\sqrt{3}\)

3, Ta có \(\dfrac{2\left(\sqrt{x}-2\right)}{x}-\dfrac{1}{4}\ge0\Leftrightarrow\dfrac{8\left(\sqrt{x}-2\right)-x}{4x}\ge0\)

\(\Rightarrow-x+8\sqrt{x}-16\ge0\Leftrightarrow-\left(\sqrt{x}-4\right)^2\ge0\)

\(\Leftrightarrow\left(\sqrt{x}-4\right)^2\le0\Leftrightarrow\sqrt{x}-4\le0\Leftrightarrow x\le16\)

Kết hợp đk vậy 0 < x =< 16 ; x khác 4