Question

Cho biểu thức M= \(\dfrac{x}{x-4}+\dfrac{\sqrt{x}}{x-2\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\) ( với x > 0, x ≠ 4)

a. Rút gọn M

b. Tìm x để M= 2

c. Tính giá trị của M khi x = 7 + \(4\sqrt{3}\)

Answer 1

\(a.M=\dfrac{x}{x-4}+\dfrac{\sqrt{x}}{x-2\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}=\dfrac{x+\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-2}\) \(b.M=2\) ⇔ \(\dfrac{\sqrt{x}}{\sqrt{x}-2}=2\text{⇔}2\sqrt{x}-4=\sqrt{x}\text{⇔}\sqrt{x}=4\text{⇔}x=16\left(TM\right)\)

\(c.x=7+4\sqrt{3}=4+2.2\sqrt{3}+3=\left(2+\sqrt{3}\right)^2\left(TM\right)\)

⇒ \(\sqrt{x}=2+\sqrt{3}\)

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