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`
