Question

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

a) Rút gọn biểu thức

b) Tính giá trị của \(\sqrt{A}\) khi x = \(4+2\sqrt{3}\)

Answer 1

điều kiện xác định : \(x\ge0;x\ne1\)

a) ta có : \(A=\left(\dfrac{2\sqrt{x}+x}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right):\left(\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\)

\(\Leftrightarrow A=\left(\dfrac{2\sqrt{x}+x}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{1}{\sqrt{x}-1}\right)\left(\dfrac{x+\sqrt{x}+1}{\sqrt{x}+2}\right)\) \(\Leftrightarrow A=\left(\dfrac{2\sqrt{x}+x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\left(\dfrac{x+\sqrt{x}+1}{\sqrt{x}+2}\right)\) \(\Leftrightarrow A=\left(\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\left(\dfrac{x+\sqrt{x}+1}{\sqrt{x}+2}\right)=\dfrac{1}{\sqrt{x}+2}\)

b) ta có : khi \(x=4+2\sqrt{3}\Rightarrow\sqrt{x}=\sqrt{4+2\sqrt{3}}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)

\(\Rightarrow\sqrt{A}=\sqrt{\dfrac{1}{\sqrt{3}+1+2}}=\sqrt{\dfrac{1}{3+\sqrt{3}}}\)

Answer 2

Mysterious Person giúp e voi