\(\dfrac{x+12}{x-4}-\dfrac{4}{\sqrt{x}-2}\left(đk:x\ne4,x\ge0\right)\\ =\dfrac{x+12}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{4\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ =\dfrac{x+12-4\sqrt{x}-8}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ =\dfrac{x-4\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ =\dfrac{\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ =\dfrac{\sqrt{x}-2}{\sqrt{x}+2}\)
Với `x >= 0,x \ne 4` có:
`[x+12]/[x-4]-4/[\sqrt{x}-2]`
`=[x+12-4(\sqrt{x}+2)]/[(\sqrt{x}-2)(\sqrt{x}+2)]`
`=[x+12-4\sqrt{x}-8]/[(\sqrt{x}-2)(\sqrt{x}+2)]`
`=[x-4\sqrt{x}+4]/[(\sqrt{x}-2)(\sqrt{x}+2)]`
`=[(\sqrt{x}-2)^2]/[(\sqrt{x}-2)(\sqrt{x}+2)]`
`=[\sqrtx{x}-2]/[\sqrt{x}+2]`
