Question

Cho biểu thức S=\(\left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{1-\sqrt{x}}{x+\sqrt{x}}\right)\) (Với x>0, x\(\ne\)1)

a. Rút gọn S

b. Tính giá trị biểu thứ S khi x=\(\dfrac{2}{2+\sqrt{3}}\)

c. Tìm x thỏa mãn P\(\sqrt{x}=2\sqrt{x}+7-\sqrt{x-4}\)

Answer 1

a: \(S=\dfrac{x+1}{\sqrt{x}}:\dfrac{x-1+1-\sqrt{x}}{x+\sqrt{x}}\)

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

b: Khi \(x=\dfrac{2}{2+\sqrt{3}}=4-2\sqrt{3}\) vào S, ta được:

\(S=\dfrac{\left(4-2\sqrt{3}+1\right)\left(\sqrt{3}-1+1\right)}{\left(\sqrt{3}-1\right)\left(4-2\sqrt{3}-1\right)}\)

\(=\dfrac{\left(5-2\sqrt{3}\right)\cdot\sqrt{3}}{\left(\sqrt{3}-1\right)\left(3-2\sqrt{3}\right)}\)