Bài 16

`a)` Ta có: `x=6+2\sqrt5=(\sqrt5+1)^2`

`->\sqrtx=\sqrt5+1(tm)`

`->A=(2(\sqrt5+1)+3)/(2(\sqrt5+1)-2)`

`=(2\sqrt5+5)/(2\sqrt5)=(2+\sqrt5)/2`

`b)B=(\sqrtx+1)/(\sqrtx+2)+(\sqrtx-2)/(1-\sqrtx)+(2x+\sqrtx-6)/(x+\sqrtx-2)(x>=0;x\ne9)`

`=(x-1-x+4+2x+\sqrtx-6)/((\sqrtx-1)(\sqrtx+2))`

`=(2x+\sqrtx-3)/((\sqrtx-1)(\sqrtx+2))`

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

`=(2\sqrtx+3)/(\sqrtx+2)`

`c)P=A.B=(2\sqrtx+3)/(2\sqrtx-2).(\sqrtx+2)/(2\sqrtx+3)=(\sqrtx+2)/(2\sqrtx-2)`

`->2P=(2\sqrtx-2+6)/(2\sqrtx-2)=1+6/(2\sqrtx-2)=1+3/(\sqrtx-1)`

Để `P` nguyên `2P` nguyên `->3/(\sqrtx-1)` nguyên

`->\sqrtx-1\in{1;-1;3}` (do `\sqrtx-1>=-1`)

`->\sqrtx\in{2;0;4}`

`->x\in{4;0;16}`

Bài 17

`a)` Ta có: `B=4/(\sqrtx+3)+(2x-\sqrtx-13)/(x-9)-\sqrtx/(\sqrtx-3)(x>=0;x\ne9)`

`=(4(\sqrtx-3)+2x-\sqrtx-13-\sqrtx(\sqrtx+3))/((\sqrtx-3)(\sqrtx+3))`

`=(4\sqrtx-12+2x-\sqrtx-13-x-3\sqrtx)/((\sqrtx-3)(\sqrtx+3))`

`=(x-25)/((\sqrtx-3)(\sqrtx+3))`

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

Khi đó:`P=B/A=((\sqrtx-5)(\sqrtx+5))/((\sqrtx-3)(\sqrtx+3)):(\sqrtx+5)/(\sqrtx-3)=(\sqrtx-5)/(\sqrtx+3)`

`b)` Ta có:`P=(\sqrtx-5)/(\sqrtx+3)=(\sqrtx+3-8)/(\sqrtx+3)=1-8/(\sqrtx+3)`

Để `P` nguyên

`->\sqrtx+3\in{4;8}` (do `\sqrtx+3>=3`)

`->\sqrtx\in{1;5}`

`->x\in{1;25}`

Mà `x` nguyên nhỏ nhất `->x=1`