b) \(P=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}+\dfrac{2+8\sqrt{x}}{x-1}-\dfrac{2}{1-\sqrt{x}}\)
\(\Rightarrow P=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\dfrac{2+8\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{2\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Rightarrow P=\dfrac{x-3\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\dfrac{2+8\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{2\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Rightarrow P=\dfrac{x-3\sqrt{x}+2+2+8\sqrt{x}+2\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(\Rightarrow P=\dfrac{x+7\sqrt{x}+6}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(\Rightarrow P=\dfrac{\left(x+\sqrt{x}\right)+\left(6\sqrt{x}+6\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(\Rightarrow P=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+6\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(\Rightarrow P=\dfrac{\sqrt{x}+6}{\sqrt{x}-1}\)
\(\Leftrightarrow Q=\dfrac{\sqrt{x}-1}{\sqrt{x}}.\dfrac{\sqrt{x}+6}{\sqrt{x}-1}+\dfrac{x-5}{\sqrt{x}}\)
\(\Leftrightarrow Q=\dfrac{\sqrt{x}+6}{\sqrt{x}}+\dfrac{x-5}{\sqrt{x}}\)
\(\Leftrightarrow Q=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}\)
\(\Leftrightarrow Q=\sqrt{x}+1+\dfrac{1}{\sqrt{x}}\)
Áp dụng BĐT Cô-si ta có:
\(Q=\sqrt{x}+\dfrac{1}{\sqrt{x}}+1\ge\dfrac{2.\sqrt{x}}{\sqrt{x}}+1=2+1=3\)
Dấu "=" xảy ra \(\Leftrightarrow\sqrt{x}=\dfrac{1}{\sqrt{x}}\Leftrightarrow x=1\)
