`A={15\sqrt{x}-11}/{x+2\sqrt{x}-3}+{3\sqrt{x}-2}/{1-\sqrt{x}}-{2\sqrt{x}+3}/{3+\sqrt{x}}`
Ta có: `x+2\sqrt{x}-3`
`=x-\sqrt{x}+3\sqrt{x}-3`
`=\sqrt{x}(\sqrt{x}-1)+3(\sqrt{x}-1)`
`=(\sqrt{x}-1)(\sqrt{x}+3)`
$\\$
$ĐKXĐ: \begin{cases}x\ge 0\\\sqrt{x}-1\ne 0\\\sqrt{x}+3\ne 0\ (luôn\ đúng\ với\ x\ge 0)\end{cases}$
`=>`$\begin{cases}x\ge 0\\x\ne 1\end{cases}$
$\\$
`a)` `A={15\sqrt{x}-11}/{x+2\sqrt{x}-3}+{3\sqrt{x}-2}/{1-\sqrt{x}}-{2\sqrt{x}+3}/{3+\sqrt{x}}`
`\qquad (x\ge 0;x\ne 1)`
`={15\sqrt{x}-11}/{(\sqrt{x}-1)(\sqrt{x}+3)}-{3\sqrt{x}-2}/{\sqrt{x}-1}-{2\sqrt{x}+3}/{\sqrt{x}+3}`
`={15\sqrt{x}-11-(3\sqrt{x}-2)(\sqrt{x}+3)-(2\sqrt{x}+3)(\sqrt{x}-1)}/{(\sqrt{x}-1)(\sqrt{x}+3)}`
`={15\sqrt{x}-11-(3x+9\sqrt{x}-2\sqrt{x}-6)-(2x-2\sqrt{x}+3\sqrt{x}-3)}/{(\sqrt{x}-1)(\sqrt{x}+3)}`
`={15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}/{(\sqrt{x}-1)(\sqrt{x}+3)}`
`={-5x+7\sqrt{x}-2}/{(sqrt{x}-1)(\sqrt{x}+3)}`
`={(\sqrt{x}-1)(-5\sqrt{x}+2)}/{(\sqrt{x}-1)(\sqrt{x}+3)}`
`={-5\sqrt{x}+2}/{\sqrt{x}+3}`
Vậy `A={-5\sqrt{x}+2}/{\sqrt{x}+3}` với `x\ge 0;x\ne 1`
$\\$
`b)` `A=1/ 2`
`<=>{-5\sqrt{x}+2}/{\sqrt{x}+3}=1/ 2`
`<=>2(-5\sqrt{x}+2)=\sqrt{x}+3`
`<=>-10\sqrt{x}+4=\sqrt{x}+3`
`<=>-11\sqrt{x}=-1`
`<=>\sqrt{x}=1/{11}`
`<=>x=(1/{11})^2=1/{121}` (thỏa mãn)
Vậy `x=1/{121}` thì `A=1/ 2`
$\\$
`c)` `A< 1/ 2`
`<=>{-5\sqrt{x}+2}/{\sqrt{x}+3}< 1/ 2`
`<=>{-5\sqrt{x}+2}/{\sqrt{x}+3}-1/ 2<0`
`<=>{2.(-5\sqrt{x}+2)-(\sqrt{x}+3)}/{2(\sqrt{x}+3)}<0`
`<=>-10\sqrt{x}+4-\sqrt{x}-3<0`
(Vì `2(\sqrt{x}+3)>0` với mọi `x\ge 0;x\ne 1)`
`<=>-11\sqrt{x}< -1`
`<=>\sqrt{x}> 1/{11}`
`<=>x> (1/{11})^2`
`<=>x>1/{121}`
Vậy `x>1/{121}` thì `A< 1/ 2`
$\\$
`d)` `A={-5\sqrt{x}+2}/{\sqrt{x}+3}`
`=>A(\sqrt{x}+3)=-5\sqrt{x}+2`
`=>A\sqrt{x}+3A+5\sqrt{x}=2`
`=>(A+5)\sqrt{x}=2-3A` $(1)$
$\\$
Nếu `A+5=0<=>A=-5`
`(1)<=>0=2-3.(-5)<=>0=17` (vô lý)
`=>A\ne -5`
`(1)<=>\sqrt{x}={2-3A}/{A+5}`
Vì `\sqrt{x}\ge 0`
`=>{2-3A}/{A+5}\ge 0`
`=>`$\left[\begin{array}{l}\begin{cases}2-3A\ge 0\\A+5>0\end{cases}\\\begin{cases}2-3A\le 0\\A+5<0\end{cases}\end{array}\right.$`=>`$\left[\begin{array}{l}\begin{cases}A\le \dfrac{2}{3}\\A> -5\end{cases}\\\begin{cases}A\ge \dfrac{2}{3}\\A< -5\end{cases}\ (loại)\end{array}\right.$`=>-5 <A\le 2/ 3`
Vì `A\in ZZ=>A\in {-4;-3;-2;-1;0}`
$\\$
+) `A=-4`
`=>{-5\sqrt{x}+2}/{\sqrt{x}+3}=-4`
`=>-5\sqrt{x}+2=-4\sqrt{x}-12`
`=>-\sqrt{x}=-14`
`=>\sqrt{x}=14=>x=14^2=196`(thỏa mãn)
$\\$
+) `A=-3`
`=>{-5\sqrt{x}+2}/{\sqrt{x}+3}=-3`
`=>-5\sqrt{x}+2=-3\sqrt{x}-9`
`=>-2\sqrt{x}=-11`
`=>\sqrt{x}={11}/2=>x={121}/4` (thỏa mãn)
$\\$
+) `A=-2`
`=>{-5\sqrt{x}+2}/{\sqrt{x}+3}=-2`
`=>-5\sqrt{x}+2=-2\sqrt{x}-6`
`=>-3\sqrt{x}=-8`
`=>\sqrt{x}=8/3=>x={64}/9`
$\\$
+) `A=-1`
`=>{-5\sqrt{x}+2}/{\sqrt{x}+3}=-1`
`=>-5\sqrt{x}+2=-\sqrt{x}-3`
`=>-4\sqrt{x}=-5`
`=>\sqrt{x}=5/4=>x={25}/{16}` (thỏa mãn)
$\\$
+) `A=0`
`=>{-5\sqrt{x}+2}/{\sqrt{x}+3}=0`
`=>-5\sqrt{x}+2=0`
`=>-5\sqrt{x}=-2`
`=>\sqrt{x}=2/ 5=>x=4/{25}` (thỏa mãn)
$\\$
Vậy `x\in {196; {121}/4; {64}/9; {25}/{16}; 4/{25}}` thì `A` có giá trị nguyên
