Đáp án:
`0<=x<9`
Giải thích các bước giải:
`A=\sqrt{4+2\sqrt3}/(\sqrt3+1)+(5+3\sqrt5)/\sqrt5-(\sqrt5+3)`
`A=\sqrt{3+2\sqrt3+1}/(\sqrt3+1)+(\sqrt5(3+\sqrt5))/\sqrt5-(3+\sqrt5)`
`A=\sqrt{(\sqrt3+1)^2}/(\sqrt3+1)+3+\sqrt5-(3+\sqrt5)`
`A=(\sqrt3+1)/(\sqrt3+1)`
`A=1`
`B=1/(3-\sqrtx)+\sqrtx/(3+\sqrtx)-(x+9)/(x-9)(x>=0,x\ne9)`
`B=-1/(\sqrt{x}-3)+\sqrtx/(\sqrtx+3)-(x+9)/(x-9)`
`B=(\sqrtx(\sqrtx-3))/((\sqrtx-3)(\sqrtx+3))-(\sqrtx+3)/((\sqrtx-3)(\sqrtx+3))-(x+9)/((\sqrtx-3)(\sqrtx+3))`
`B=(x-3\sqrtx-\sqrtx-3-x-9)/((\sqrtx-3)(\sqrtx+3))`
`B=(-4\sqrtx-12)/((\sqrtx-3)(\sqrtx+3))`
`B=(-4(\sqrtx+3))/((\sqrtx-3)(\sqrtx+3))`
`B=-4/(\sqrtx-3)`
`B>A`
`<=>-4/(\sqrtx-3)>1`
`<=>-4/(\sqrtx-3)-1>0`
`<=>(-4-\sqrtx+3)/(\sqrtx-3)>0`
`<=>(-\sqrtx-1)/(\sqrtx-3)>0`
Mà `-\sqrtx-1<=-1<0`
`<=>\sqrtx-3<0`
`<=>\sqrtx<3<=>0<=x<9`