a,
`A={3x}/{x\sqrt{x}+1}-{\sqrt{x}-1}/{x-\sqrt{x}+1}-1/{1+\sqrt{x}}`
`={3x}/{(\sqrt{x}+1)(x-\sqrt{x}+1)}-{(\sqrt{x}-1)(\sqrt{x}+1)}/{(\sqrt{x}+1)(x-\sqrt{x}+1)}-{x-\sqrt{x}+1}/{(\sqrt{x}+1)(x-\sqrt{x}+1)}`
`={3x-x+1-x+\sqrt{x}-1}/{(\sqrt{x}+1)(x-\sqrt{x}+1)}={x+\sqrt{x}}/{(\sqrt{x}+1)(x-\sqrt{x}+1)}`
`={\sqrt{x}(\sqrt{x}+1)}/{(\sqrt{x}+1)(x-\sqrt{x}+1)}=\sqrt{x}/{x-\sqrt{x}+1}`
b,
Ta có:
`A-1=\sqrt{x}/{x-\sqrt{x}+1}-1={-x+2\sqrt{x}-1}/{x-\sqrt{x}+1}={-(\sqrt{x}-1)^2}/{x-\sqrt{x}+1}\le 0`
`=>A\le 1`. Đẳng thức xảy ra `<=>\sqrt{x}-1=0<=>x=1`
Vậy `max A=1` đạt được khi `x=1`
