1
ĐK: \(x\ge0;x\ne4\)
Biểu thức trở thành:
\(\left(\dfrac{\sqrt{x}+2}{2\left(\sqrt{x}-2\right)}-\dfrac{4\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\left(\dfrac{\sqrt{x}}{2}-\dfrac{2}{2}\right)\)
\(=\left(\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+2\right)}{2\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{4\sqrt{x}.2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right).2}\right):\left(\dfrac{\sqrt{x}-2}{2}\right)\)
\(=\left(\dfrac{\left(\sqrt{x}+2\right)^2}{2\left(x-4\right)}-\dfrac{8\sqrt{x}}{2\left(x-4\right)}\right).\left(\dfrac{2}{\sqrt{x}-2}\right)\\ =\left(\dfrac{x+4\sqrt{x}+4-8\sqrt{x}}{2\left(x-4\right)}\right).\left(\dfrac{2}{\sqrt{x}-2}\right)\\ =\dfrac{x-4\sqrt{x}+4}{2\left(x-4\right)}.\dfrac{2}{\sqrt{x}-2}\\ =\dfrac{\left(\sqrt{x}-2\right)^2.2}{2\left(x-4\right)\left(\sqrt{x}-2\right)}\\ =\dfrac{\sqrt{x}-2}{x-4}\)
2
ĐK: \(x\ge0;x\ne1\)
Biểu thức trở thành:
\(\left(\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x-1}-\dfrac{5\sqrt{x}-3}{x-1}\right)\left(\dfrac{2\left(\sqrt{x}+3\right)}{\sqrt{x}+3}-\dfrac{4}{\sqrt{x}+3}\right)\\ =\left(\dfrac{x+\sqrt{x}-5\sqrt{x}+3}{x-1}\right)\left(\dfrac{2\sqrt{x}+6-4}{\sqrt{x}+3}\right)\\ =\dfrac{x-4\sqrt{x}+3}{x-1}.\dfrac{2\sqrt{x}+2}{\sqrt{x}+3}\\ =\dfrac{x-3\sqrt{x}-\sqrt{x}+3}{x-1}.\dfrac{2\sqrt{x}+2}{\sqrt{x}+3}\\ =\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)-\left(\sqrt{x}-3\right)}{x-1}.\dfrac{2\left(\sqrt{x}+1\right)}{\sqrt{x}+3}\\ =\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right).2.\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{2\sqrt{x}-6}{\sqrt{x}+3}\)
3
\(x>0;x\ne4;x\ne1\)
Biểu thức trở thành:
\(\left(\dfrac{2\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)^2}-\dfrac{\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right).-\left(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right)\)
\(=\left(\dfrac{2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)}\right).-\sqrt{x}\)
\(=\dfrac{2\sqrt{x}-\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}-2\right)}.-\sqrt{x}\\ =-\dfrac{\left(\sqrt{x}-2\right).\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\\ =-1\)
4
ĐK: \(x>0;x\ne1\)
Biểu thức trở thành:
\(\dfrac{2}{3\sqrt{x}}-\dfrac{1}{2\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}-4}{6\sqrt{x}\left(1-\sqrt{x}\right)}\\ =\dfrac{2}{3\sqrt{x}}-\dfrac{1}{2\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}-4}{6\sqrt{x}\left(\sqrt{x}-1\right)}\\ =\dfrac{2.2.\left(\sqrt{x}-1\right)}{6\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}.3}{6\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}-4}{6\sqrt{x}\left(\sqrt{x}-1\right)}\\ =\dfrac{4\sqrt{x}-4-3\sqrt{x}+\sqrt{x}-4}{6x-6\sqrt{x}}\\ =\dfrac{2\sqrt{x}-8}{6x-6\sqrt{x}}\)
