Đáp án:
`a)` `x>0;x\ne 4`; `P=-\sqrt{x}+1`
`b)` `P=2-\sqrt{2}`
`c)` `max\ P\sqrt{x}=1/ 4` khi `x=1/ 4`
Giải thích các bước giải:
`a)` `P=({\sqrt{x}-4}/{x-2\sqrt{x}}+3/{\sqrt{x}-2}):({\sqrt{x}+2}/{\sqrt{x}}-{\sqrt{x}}/{\sqrt{x}-2})`
`ĐKXĐ: x> 0; \sqrt{x}-2\ne 0`
`=>x>0; x\ne 4`
`P={\sqrt{x}-4+3.\sqrt{x}}/{\sqrt{x}.(\sqrt{x}-2)}:{(\sqrt{x}+2).(\sqrt{x}-2)-x}/{\sqrt{x}(\sqrt{x}-2)}`
`={4\sqrt{x}-4}/{\sqrt{x}.(\sqrt{x}-2)}:{x-4-x}/{\sqrt{x}(\sqrt{x}-2)}`
`={-4.(-\sqrt{x}+1)}/{\sqrt{x}.(\sqrt{x}-2)}.{\sqrt{x}(\sqrt{x}-2)}/{-4}`
`=-\sqrt{x}+1`
Vậy `P=-\sqrt{x}+1`
$\\$
`b)` Với `x=3-2\sqrt{2}` (thỏa đk)
`=>x=2-2\sqrt{2}+1=(\sqrt{2}-1)^2`
Ta có:
`P=-\sqrt{x}+1`
`=-\sqrt{(\sqrt{2}-1)^2}+1`
`=-|\sqrt{2}-1|+1`
`=-(\sqrt{2}-1)+1=2-\sqrt{2}`
Vậy `P=2-\sqrt{2}` khi `x=3-2\sqrt{2}`
$\\$
`c)` Ta có:
`P\sqrt{x}=(-\sqrt{x}+1).\sqrt{x}` $(x>0;x\ne 4)$
`=-x+\sqrt{x}`
`=-(x-2.\sqrt{x} . 1/ 2 +1/ 4)+1/ 4`
`=-(\sqrt{x}-1/ 2)^2+1/ 4`
Với mọi `x>0;x\ne 4` ta có:
`\qquad (\sqrt{x}-1/ 2)^2\ge 0`
`=>-(\sqrt{x}-1/ 2)^2\le 0`
`=>-(\sqrt{x}-1/ 2)^2+1/ 4\le 1/ 4`
`=>P\sqrt{x}\le 1/ 4`
Dấu "=" xảy ra khi `(\sqrt{x}-1/ 2)^2=0`
`<=>\sqrt{x}=1/ 2 <=>x=1/ 4`
Vậy `P\sqrt{x}` có $GTLN$ bằng `1/ 4` khi `x= 1/ 4`
