Question

1. Cho biểu thức: A = \(\left(\dfrac{x-5\sqrt{x}}{x-5}-1\right):\left(\dfrac{25-x}{x+2\sqrt{x}-15}-\dfrac{\sqrt{x}+3}{\sqrt{x}+5}+\dfrac{\sqrt{x}-5}{\sqrt{x}-3}\right).\)

a) Rút gọn A.

b) Tìm x để A \(< 1.\) \(\)

Answer 1

\(a.A=\left(\dfrac{x-5\sqrt{x}}{x-25}-1\right):\left(\dfrac{25-x}{x+2\sqrt{x}-15}-\dfrac{\sqrt{x}+3}{\sqrt{x}+5}+\dfrac{\sqrt{x}-5}{\sqrt{x}-3}\right)=\dfrac{-5}{\sqrt{x}+5}:\dfrac{25-x-x+9+x-25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}=\dfrac{-5}{\sqrt{x}+5}.\dfrac{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}{-\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{5}{\sqrt{x}+3}\) ( x ≥0 ; x # 9 ; x # 25 )

\(b.A< 1\) ⇔ \(\dfrac{5}{\sqrt{x}+3}< 1\)

⇔ \(\dfrac{2-\sqrt{x}}{\sqrt{x}+3}< 0\)

⇔ \(2-\sqrt{x}< 0\)

⇔ \(x>4\) ( x # 9 ; x # 25 )

KL.................