Question

cho E = \(\left(\dfrac{3+\sqrt{x}}{x-1}+\dfrac{3}{\sqrt{x}+1}\right):\dfrac{4}{x+\sqrt{x}}\)

a. Tìm ĐK và rút gọn E

b. Tính E khi x = \(\dfrac{9}{4}\)

c. Tìm x để E < 0

Answer 1

\(a.E=\left(\dfrac{3+\sqrt{x}}{x-1}+\dfrac{3}{\sqrt{x}+1}\right):\dfrac{4}{x+\sqrt{x}}=\dfrac{4\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{4}=\dfrac{x}{\sqrt{x}-1}\) ( x ≥ 0 ; x # 1 )

\(b.x=\dfrac{9}{4}\left(TMĐKXĐ\right)\) ⇒ \(\sqrt{x}=\dfrac{3}{2}\) , ta có :

\(E=\dfrac{9}{4}:\left(\dfrac{3}{2}-1\right)=\dfrac{9}{4}.2=\dfrac{9}{2}\)

\(c.E< 0\) ⇔ \(\dfrac{x}{\sqrt{x}-1}< 0\)

⇔ \(\sqrt{x}-1< 0\) ⇔ \(x< 1\)

Kết hợp vs ĐKXĐ : \(0\text{≤}x< 1\)

KL.......