Question

Cho biểu thức
A=\(\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{1}{x+\sqrt{x}}\right):\dfrac{x-\sqrt{x}+1}{x\sqrt{x}+1}\)
a,Rút gọn biểu thức
b,Tìm x để A<0

Answer 1

a , Ta có :

\(A=\left[\dfrac{1}{\sqrt{x}+1}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right].\dfrac{\sqrt{x^3}+1}{x-\sqrt{x}+1}\)

\(A=\left[\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right].\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}\)

\(A=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

\(A=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)

b , \(A< 0\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}}< 0\Leftrightarrow\sqrt{x}-1< 0\Leftrightarrow\sqrt{x}< 1\Leftrightarrow x< 1\)