a)Đkxđ : x#1 , x > 0
Q = \(\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{x-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{x-1}\right)\)
Q = \(\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
Q=\(\left(\dfrac{x}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
Q=\(\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}-1+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
Q=\(\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}X\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\)
Q=\(\dfrac{x-1}{\sqrt{x}}\)
b)Thay x = 2\(\sqrt{2}\)+3 vào phương trình ta được :
Q=\(\dfrac{2\sqrt{2}+3-1}{\sqrt{2\sqrt{2}+3}}\)
Q=\(\dfrac{2\sqrt{2}+2}{\sqrt{\left(\sqrt{2}+1\right)}^2}\)
Q=\(\dfrac{2\left(\sqrt{2}+1\right)}{\sqrt{2}+1}\)
Q= 2
