Đáp án:

Chuẩn hóa:`x\in ZZ`(câu c)

Giải thích các bước giải:

`1)P=((2sqrtx)/(sqrtx-3)-(sqrtx+2)/(sqrtx+3)+(5sqrtx+6)/(9-x)):(1-3/(sqrtx+3))(x>0,x ne 9)`

`P=((2sqrtx(sqrtx+3))/((sqrtx-3)(sqrtx+3))-((sqrtx+2)(sqrtx-3))/((sqrtx-3)(sqrtx+3))-(5sqrtx+6)/((sqrtx-3)(sqrtx+3))):((sqrtx+3-3)/(sqrtx+3))`

`P=((2x+6sqrtx-(x-sqrtx-6)-5sqrtx-6)/((sqrtx-3)(sqrtx+3))):sqrtx/(sqrtx+3)`

`P=(x+2sqrtx)/((sqrtx-3)(sqrtx+3))*(sqrtx+3)/sqrtx`

`P=(sqrtx(sqrtx+2))/((sqrtx-3)(sqrtx+3))*(sqrtx+3)/sqrtx`

`P=(sqrtx+2)/(sqrtx-3)`

`x=2/(2-sqrt3)`

`x=(2(2+sqrt3))/(4-3)=4+2sqrt3`

`x=3+2sqrt3+1=(sqrt3+1)^2`

`=>sqrtx=sqrt3+1`

`=>P=(sqrt3+1+2)/(sqrt3+1-3)`

`=>P=(3+sqrt3)/(sqrt3-2)`

`=>P=((3+sqrt3)(2+sqrt3))/(3-4)`

`=>P=-(6+3sqrt3+2sqrt3+3)`

`=>P=-(9+5sqrt3)`

`2)A<1`

`<=>(sqrtx+2)/(sqrtx-3)-1<0`

`<=>(sqrtx+2-sqrtx+3)/(sqrtx-3)<0`

`<=>5/(sqrtx-3)<0`

Mà `5>0`

`<=>sqrtx-3<0`

`<=>sqrtx<3`

`<=>x<9`

`=>0<x<9`

`3)1/P=(sqrtx-3)/(sqrtx+2)`

`P=(sqrtx+2-5)/(sqrtx+2)`

`P=1-5/(sqrtx+2)`

`x>0,x in ZZ`

`=>x>=1`

`=>sqrtx>=1`

`=>sqrtx+2>=3>0`

`=>5/(sqrtx+3)<=5/3`

`=>P>=1-5/3=-2/3`

Dấu "=" xảy ra khi `x=1`

`4)M=sqrtx+P(x>9)`

`M=sqrtx+(sqrtx+2)/(sqrtx-3)`

`M=sqrtx+(sqrtx-3+5)/(sqrtx-3)`

`M=sqrtx+1+5/(sqrtx-3)`

`M=sqrtx-3+5/(sqrtx-3)+4`

Áp dụng bđt cosi

`=>sqrtx-3+5/(sqrtx-3)>=2sqrt5`

`=>M>=4+2sqrt5`

Dấu "=" xảy ra khi `sqrtx-3=5/(sqrtx-3)`

`<=>(sqrtx-3)^2=5`

`<=>sqrtx-3=sqrt5` (do `x>9=>sqrtx>3=>sqrtx-3>0`)

`<=>sqrtx=3+sqrt5`

`<=>x=9+5+6sqrt5=14+6sqrt5`

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