Đáp án+Giải thích các bước giải:
`a)A=(sqrtx-3)/(sqrtx+3)+(sqrtx+3)/(sqrtx-3)-(12sqrtx)/(x-9)(x>=0,x ne 9)`
`=((sqrtx-3)^2+(sqrtx+3)^2-12sqrtx)/(x-9)`
`=(x-6sqrtx+9+x+6sqrtx+9-12sqrtx)/(x-9)`
`=(2(x-6sqrtx+9))/(x-9)`
`=(2(sqrtx-3)^2)/((sqrtx-3)(sqrtx+3))`
`=(2(sqrtx-3))/(sqrtx+3).`
b)A=1`
`<=>2(sqrtx-3)=sqrtx+3`
`<=>2sqrtx-6=sqrtx+3``
`<=>sqrtx=9`
`<=>x=81(tmđk)`
`c)A=(2(sqrtx+3-6))/(sqrtx+3)`
`=2-12/(sqrtx+3)`
`x>=0=>sqrtx+3>=3>0`
`=>12/(sqrtx+3)<=12/3=4`
`=A>=2-4=-2`
Dấu "=" xảy ra khi `x=0`
`d)A in ZZ`
`=>2(sqrtx-3) vdots sqrtx+3`
`=>2(sqrtx+3-6) vdots sqrtx+3`
`=>12 vdots sqrtx+3`
`=>sqrtx+3 in Ư(12)={+-1,+-2,+-3,+-4,+-6,+-12}`
Mà `sqrtx+3>=3`
`=>sqrtx+3 in {3,4,6,12}`
`=>sqrtx in {0,1,3,9}`
`=>x in {0,1,9,81}`
