Question

Cho biểu thức:

\(B=\left(1+\dfrac{\sqrt{x}}{x+1}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2\sqrt{x}}{x\sqrt{x}+\sqrt{x}-x-1}\right)\)

a. Rút gọn B

b. Tính B khi \(x=3+2\sqrt{2}\)

c.Giải phương trình \(B=\sqrt{5}\)

Answer 1

\(a.B=\left(1+\dfrac{\sqrt{x}}{x+1}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2\sqrt{x}}{x\sqrt{x}+\sqrt{x}-x-1}\right)=\dfrac{x+\sqrt{x}+1}{x+1}:\dfrac{x+1-2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}=\dfrac{x+\sqrt{x}+1}{x+1}.\dfrac{\left(x+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)^2}=\dfrac{x+\sqrt{x}+1}{\sqrt{x}-1}\) ( x ≥ 0 ; x # 1 )

\(b.x=3+2\sqrt{2}=2+2\sqrt{2}+1=\left(\sqrt{2}+1\right)^2\left(TM\right)\)

⇒ \(\sqrt{x}=\sqrt{2}+1\)

Khi đó : \(B=\dfrac{3+2\sqrt{2}+\sqrt{2}+1+1}{\sqrt{2}+1-1}=\dfrac{5+3\sqrt{2}}{\sqrt{2}}\)

\(c.B=\sqrt{5}\text{⇔}\dfrac{x+\sqrt{x}+1}{\sqrt{x}-1}-\sqrt{5}=0\)

\(\text{⇔}x+\sqrt{x}+1-\sqrt{5x}+\sqrt{5}=0\)

Tự giải ra nhé :D ( @.@ )