\(1,đkx\ne1;x>0\\ =\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\\ =\dfrac{\sqrt{x}.\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\\ =\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\\ ---------------------------\\ \)
\(2,đkx>0;x\ne3\\ =\dfrac{\sqrt{x}-3}{\sqrt{x}}-\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{\sqrt{x}+9}{\sqrt{x}\left(\sqrt{x}-3\right)}\\ =\dfrac{\left(\sqrt{x}-3\right)^2-x-\sqrt{x}-9}{\sqrt{x}\left(\sqrt{x}-3\right)}\\ =\dfrac{x-6\sqrt{x}+9-x-\sqrt{x}-9}{\sqrt{x}\left(\sqrt{x}-3\right)}\\ =\dfrac{-7\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-3\right)}\\ =\dfrac{-7}{\sqrt{x}-3}\\ -----------------------\)
\(3,đkx\ne5;x>0\\ =\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)-4\left(\sqrt{x}-5\right)+\sqrt{x}-26}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+2\right)}\\ =\dfrac{x+2\sqrt{x}-4\sqrt{x}+20+\sqrt{x}-26}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+2\right)}\\ =\dfrac{x-\sqrt{x}-6}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+2\right)}=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}-3}{\sqrt{x}+5}\\ ----------------------------------\)
\(4,đkx\ne1;x>0\\ =\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{2\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\\ =\dfrac{x-2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\\ =\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}\left(\sqrt{x}-1\right)}\\ =\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
