Question

Answer 1

a: \(Q=\dfrac{3x+\sqrt{9x}-3}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{\sqrt{x}-2}{1-\sqrt{x}}\)

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

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

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

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

b: Khi \(x=4+2\sqrt{3}=\left(\sqrt{3}+1\right)^2\) thì

\(Q=\dfrac{\sqrt{\left(\sqrt{3}+1\right)^2}+1}{\sqrt{\left(\sqrt{3}+1\right)^2}-1}\)

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

Answer 2

a) \(Q=\dfrac{3x+\sqrt{9x}-3}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{\sqrt{x}-2}{1-\sqrt{x}}\)

\(Q=\dfrac{3x+3\sqrt{x}-3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\)

\(Q=\dfrac{3x+3\sqrt{x}-3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\dfrac{x-1}{\left(\sqrt{x}+2\right)\left(\sqrt{x-1}\right)}-\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

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

\(Q=\dfrac{x+3\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

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

\(Q=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)

b) Ta có:

\(x=4+2\sqrt{3}=\left(\sqrt{3}\right)^2+2\cdot\sqrt{3}\cdot1+1^2=\left(\sqrt{3}+1\right)^2\)

Thay vào Q ta có:

\(Q=\dfrac{\sqrt{\left(\sqrt{3}+1\right)^2}+1}{\sqrt{\left(\sqrt{3}+1\right)^2}-1}=\dfrac{\sqrt{3}+1+1}{\sqrt{3}+1-1}=\dfrac{\sqrt{3}+2}{\sqrt{3}}=\dfrac{\sqrt{3}\left(\sqrt{3}+2\right)}{\sqrt{3}\cdot\sqrt{3}}=\dfrac{3+2\sqrt{3}}{3}\)