giải hộ mk câu 2 nha
2. ĐKXĐ: \(x\ge0,x\ne1\)
\(P=\left(\sqrt{x}-\dfrac{x+2}{\sqrt{x}+1}\right):\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{\sqrt{x}-4}{1-x}\right)\)
\(=\dfrac{x+\sqrt{x}-x-2}{\sqrt{x}+1}:\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}:\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)+\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}:\dfrac{x-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}.\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\)
\(P=\dfrac{1}{2}\Rightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}+2}=\dfrac{1}{2}\Rightarrow2\sqrt{x}-2=\sqrt{x}+2\Rightarrow\sqrt{x}=4\Rightarrow x=16\)
b) Ta có: \(P=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}=1-\dfrac{3}{\sqrt{x}+2}\)
Ta có: \(\sqrt{x}+2\ge2\Rightarrow\dfrac{3}{\sqrt{x}+2}\le\dfrac{3}{2}\Rightarrow1-\dfrac{3}{\sqrt{x}+2}\ge-\dfrac{1}{2}\)
\(\Rightarrow P_{min}=-\dfrac{1}{2}\) khi \(x=0\)
