Đáp án:
`((\sqrt{x}+1)/(\sqrt{x}-2)-(2\sqrt{x})/(\sqrt{x}+2)+(5\sqrt{x}+2)/(4-x)):(3\sqrt{x}-x)/(x+4\sqrt{x}+4)=(\sqrt{x}+2)/(\sqrt{x}-3)`
Giải thích các bước giải:
`ĐKXĐ: x>0; x\ne4; x\ne9`
`((\sqrt{x}+1)/(\sqrt{x}-2)-(2\sqrt{x})/(\sqrt{x}+2)+(5\sqrt{x}+2)/(4-x)):(3\sqrt{x}-x)/(x+4\sqrt{x}+4)`
`=((\sqrt{x}+1)/(\sqrt{x}-2)-(2\sqrt{x})/(\sqrt{x}+2)-(5\sqrt{x}+2)/(x-4)):(\sqrt{x}(3-\sqrt{x}))/((\sqrt{x}+2)^3)`
`=[(\sqrt{x}+1)/(\sqrt{x}-2)-(2\sqrt{x})/(\sqrt{x}+2)-(5\sqrt{x}+2)/((\sqrt{x}+2)(\sqrt{x}-2))].((\sqrt{x}+2)^2)/(\sqrt{x}(3-\sqrt{x}))`
`=((\sqrt{x}+1)(\sqrt{x}+2)-2\sqrt{x}(\sqrt{x}-2)-(5\sqrt{x}+2))/((\sqrt{x}+2)(\sqrt{x}-2)).((\sqrt{x}+2)^2)/(\sqrt{x}(3-\sqrt{x}))`
`=(x+2\sqrt{x}+\sqrt{x}+2-2x+4\sqrt{x}-5\sqrt{x}-2)/((\sqrt{x}+2)(\sqrt{x}-2)).((\sqrt{x}+2)^2)/(\sqrt{x}(3-\sqrt{x}))`
`=(-x+2\sqrt{x})/((\sqrt{x}+2)(\sqrt{x}-2)).((\sqrt{x}+2)^2)/(\sqrt{x}(3-\sqrt{x}))`
`=(-\sqrt{x}(\sqrt{x}-2))/((\sqrt{x}+2)(\sqrt{x}-2)).((\sqrt{x}+2)^2)/(\sqrt{x}(3-\sqrt{x}))`
`=(-(\sqrt{x}+2))/(3-\sqrt{x})`
`=(\sqrt{x}+2)/(\sqrt{x}-3)`
Vậy với `x>0; x\ne4; x\ne9` thì `((\sqrt{x}+1)/(\sqrt{x}-2)-(2\sqrt{x})/(\sqrt{x}+2)+(5\sqrt{x}+2)/(4-x)):(3\sqrt{x}-x)/(x+4\sqrt{x}+4)=(\sqrt{x}+2)/(\sqrt{x}-3)`
