\(\)Bài 1: Chứng minh rằng \(\left(\dfrac{1}{\sqrt{x}+2}-\dfrac{1}{\sqrt{x}+7}\right)\): \(\dfrac{5}{\sqrt{x}+7}\) = \(\dfrac{1}{\sqrt{x}+2}\)
Bài 2: Cho P= \(\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\right)\): \(\dfrac{2\sqrt{x}+1}{2x-2}\) với x ≥ 0 và x ≠ 1
a) Rút gọn P
b) Tìm x để P= \(-\dfrac{1}{2}\)
c) Tìm x để P nhận giá trị âm
1.
\(\left(\dfrac{1}{\sqrt{x}+2}-\dfrac{1}{\sqrt{x+7}}\right):\dfrac{5}{\sqrt{x}+7}=\dfrac{5}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+7\right)}.\dfrac{\sqrt{x}+7}{5}\)
\(=\dfrac{1}{\sqrt{x}+2}\)
2.
\(P=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\right):\dfrac{2\sqrt{x}+1}{2x-2}\)
\(=\left(\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\dfrac{2\left(x-1\right)}{2\sqrt{x}+1}\)
\(=\dfrac{-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{2\sqrt{x}+1}\)
\(=\dfrac{-4}{2\sqrt{x}+1}\)
b.
Để \(P=-\dfrac{1}{2}\Rightarrow\dfrac{-4}{2\sqrt{x}+1}=-\dfrac{1}{2}\)
\(\Rightarrow2\sqrt{x}+1=8\Rightarrow\sqrt{x}=\dfrac{7}{2}\)
\(\Rightarrow x=\dfrac{49}{4}\)
c.
Để \(P< 0\Rightarrow\dfrac{-4}{2\sqrt{x}+1}< 0\Rightarrow2\sqrt{x}+1>0\)
\(\Rightarrow\sqrt{x}>-\dfrac{1}{2}\) (luôn đúng)
Vậy với \(x\ge0;x\ne1\) thì P âm
