Bài 1: \(\left(\dfrac{1}{\sqrt{x}+2}-\dfrac{1}{\sqrt{x}+7}\right)\): \(\dfrac{5}{\sqrt{x}+7}\)= \(\dfrac{1}{\sqrt{x}+2}\)
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 , x ≠ 1
a) Rút gọn
b) Tìm x để P= \(-\dfrac{1}{2}\)
c) Tìm x để P nhận giá trị âm
Bài 2:
a: \(P=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\right):\dfrac{2\sqrt{x}+1}{2x-2}\)
\(=\dfrac{\sqrt{x}-1-\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{2\left(x-1\right)}{2\sqrt{x}+1}\)
\(=\dfrac{-2\cdot2}{2\sqrt{x}+1}=\dfrac{-4}{2\sqrt{x}+1}\)
b: \(P=-\dfrac{1}{2}\)
=>\(\dfrac{-4}{2\sqrt{x}+1}=\dfrac{-1}{2}\)
=>\(2\sqrt{x}+1=8\)
=>\(2\sqrt{x}=7\)
=>\(\sqrt{x}=\dfrac{7}{2}\)
=>\(x=\dfrac{49}{4}\left(nhận\right)\)
c: Để P<0 thì \(-\dfrac{4}{2\sqrt{x}+1}< 0\)
=>\(2\sqrt{x}+1>0\)(luôn đúng với mọi x thỏa mãn ĐKXĐ)
Bài 1:
\(\left(\dfrac{1}{\sqrt{x}+2}-\dfrac{1}{\sqrt{x}+7}\right):\dfrac{5}{\sqrt{x}+7}\)
\(=\dfrac{\sqrt{x}+7-\sqrt{x}-2}{\left(\sqrt{x}+7\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+7}{5}\)
\(=\dfrac{5}{5\left(\sqrt{x}+2\right)}=\dfrac{1}{\sqrt{x}+2}\)
