Đáp án:
`a)` `P={3\sqrt{x}-5}/{2\sqrt{x}+1}`
`b)` `P={3\sqrt{2}-8}/{2\sqrt{2}-1}`
`c)` `P< 3/ 2`
Giải thích các bước giải:
`a)` Ta có:
`\qquad (2\sqrt{x}-3).(\sqrt{x}+1)`
`=2x+2\sqrt{x}-3\sqrt{x}-3`
`=2x-\sqrt{x}-3`
`P=(2-{\sqrt{x}-1}/{2\sqrt{x}-3}): ({6\sqrt{x}+1}/{2x-\sqrt{x}-3}+{\sqrt{x}}/{\sqrt{x}+1})`
$ĐK: \begin{cases}x\ge 0\\2\sqrt{x}-3\ne 0\end{cases}$`<=>`$\begin{cases}x\ge 0\\\sqrt{x}\ne \dfrac{3}{2}\end{cases}$`<=>`$\begin{cases}x\ge 0\\x\ne \dfrac{9}{4}\end{cases}$
`P={2(2\sqrt{x}-3)-(\sqrt{x}-1)}/{2\sqrt{x}-3} : {6\sqrt{x}+1+\sqrt{x}. (2\sqrt{x}-3)}/{(2\sqrt{x}+3).(\sqrt{x}+1)}`
`={3\sqrt{x}-5}/{2\sqrt{x}-3} : {6\sqrt{x}+1+2x-3\sqrt{x}}/{(2\sqrt{x}-3)(\sqrt{x}+1)}`
`={3\sqrt{x}-5}/{2\sqrt{x}-3} . {(2\sqrt{x}+3).(\sqrt{x}+1)}/ {2x+3\sqrt{x}+1}`
`={(3\sqrt{x}-5).(\sqrt{x}+1)}/{2x+2\sqrt{x}+\sqrt{x}+1}`
`={(3+\sqrt{x}-5).(\sqrt{x}+1)}/{(\sqrt{x}+1).(2\sqrt{x}+1)}`
`={3\sqrt{x}-5}/{2\sqrt{x}+1}`
Vậy `P={3\sqrt{x}-5}/{2\sqrt{x}+1}`
$\\$
`b)` Với `x=3-2\sqrt{2}`
`=>x=2-2\sqrt{2}.1+1=(\sqrt{2}-1)^2`
`=>\sqrt{x}=\sqrt{2}-1`
`=>P={3\sqrt{x}-5}/{2\sqrt{x}+1}`
`={3.(\sqrt{2}-1)-5}/{2.(\sqrt{2}-1)+1}`
`={3\sqrt{2}-8}/{2\sqrt{2}-1}`
Vậy `P={3\sqrt{2}-8}/{2\sqrt{2}-1}` khi `x=3-2\sqrt{2}`
$\\$
`c)` Ta có:
`\qquad P-3/ 2 `
`={3\sqrt{x}-5}/{2\sqrt{x}+1}-3/ 2`
`={2.(3\sqrt{x}-5)-3(2\sqrt{x}+1)}/{2.(2\sqrt{x}+1)}`
`={6\sqrt{x}-10-6\sqrt{x}-3}/{4\sqrt{x}+2}`
`={-13}/{4\sqrt{x}+2}`
Với mọi `x\ge 0;x\ne 9/ 4` ta có:
`\qquad \sqrt{x}\ge 0`
`=>4\sqrt{x}+2\ge 2>0`
`=>{-13}/{4\sqrt{x}+2}<0`
`=>P-3/ 2<0=>P<3/ 2`
Vậy `P< 3/ 2`
