a) \(đk:x\ge0,x\ne1,x\ne25\)
\(P=\dfrac{2+\sqrt{x}+5+2\sqrt{x}-10}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}.\dfrac{\sqrt{x}+5}{\sqrt{x}-1}\)
\(=\dfrac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}-1\right)}=\dfrac{3}{\sqrt{x}-5}\)
b) \(P=\dfrac{3}{\sqrt{x}-5}< 0\)
\(\Leftrightarrow\sqrt{x}-5< 0\Leftrightarrow\sqrt{x}< 5\)
Kết hợp đk
\(\Leftrightarrow0\le x< 25\) và \(x\ne1\)
c) \(P=\dfrac{3}{\sqrt{x}-5}=\dfrac{3}{\sqrt{4-2\sqrt{3}}-5}=\dfrac{3}{\sqrt{\left(\sqrt{3}-1\right)^2}-5}=\dfrac{3}{\sqrt{3}-6}=\dfrac{3\sqrt{3}+18}{3}=\sqrt{3}+6\)
a: Ta có: \(P=\left(\dfrac{2}{x-25}+\dfrac{1}{\sqrt{x}-5}+\dfrac{2}{\sqrt{x}+5}\right):\dfrac{\sqrt{x}-1}{\sqrt{x}+5}\)
\(=\dfrac{2+\sqrt{x}+5+2\sqrt{x}-10}{\left(\sqrt{x}-5\right)}\cdot\dfrac{1}{\sqrt{x}-1}\)
\(=\dfrac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}-1\right)}=\dfrac{3}{\sqrt{x}-5}\)
