Bài 3: `ĐKXĐ:x\ge0,x\ne1,x\ne9`
`a,Q=({2x+1}/{x\sqrt{x}-1}-{1}/{\sqrt{x}-1}):(1-{x+4}/{x+\sqrt{x}+1})`
`=[{2x+1}/{(\sqrt{x}-1)(x+\sqrt{x}+1)}-{1}/{\sqrt{x}-1}]:(1-{x+4}/{x+\sqrt{x}+1})`
`=[{2x+1}/{(\sqrt{x}-1)(x+\sqrt{x}+1)}-{1}/{\sqrt{x}-1}]:(1-{x+4}/{x+\sqrt{x}+1})`
`=[{2x+1}/{(\sqrt{x}-1)(x+\sqrt{x}+1)}-{x+\sqrt{x}+1}/{(\sqrt{x}-1)(x+\sqrt{x}+1)}]:({x+\sqrt{x}+1}/{x+\sqrt{x}+1}-{x+4}/{x+\sqrt{x}+1})`
`={2x+1-(x+\sqrt{x}+1)}/{(\sqrt{x}-1)(x+\sqrt{x}+1)}:{x+\sqrt{x}+1-(x+4)}/{x+\sqrt{x}+1}`
`={2x+1-x-\sqrt{x}-1}/{(\sqrt{x}-1)(x+\sqrt{x}+1)}:{x+\sqrt{x}+1-x-4}/{x+\sqrt{x}+1}`
`={x-\sqrt{x}}/{(\sqrt{x}-1)(x+\sqrt{x}+1)}:{\sqrt{x}-3}/{x+\sqrt{x}+1}`
`={\sqrt{x}(\sqrt{x}-1)}/{(\sqrt{x}-1)(x+\sqrt{x}+1)}.{x+\sqrt{x}+1}/{\sqrt{x}-3}`
`={\sqrt{x}}/{\sqrt{x}-3}`
Vậy với `x\ge0,x\ne1` thì `Q={\sqrt{x}}/{\sqrt{x}-3}`
`b,Q={\sqrt{x}}/{\sqrt{x}-3}`
`={\sqrt{x}-3+3}/{\sqrt{x}-3}`
`={\sqrt{x}-3}/{\sqrt{x}-3}+{3}/{\sqrt{x}-3}`
`=1+{3}/{\sqrt{x}-3}`
`Q\inZ⇔{3}/{\sqrt{x}-3}\inZ`
`⇔\sqrt{x}-3\inƯ(3)={-3;-1;1;3}`
`⇔\sqrt{x}\in{0;2;4;6}`
`⇔x\in{0;4;16;36}(TM)`
Vậy với `x\in{0;4;16;36}` thì `Q\inZ`
Bài 4: `ĐKXĐ:x>0`
`a,P=({1}/{x+\sqrt{x}}-{1}/{\sqrt{x}+1}):{\sqrt{x}}/{x+2\sqrt{x}+1}`
`=[{1}/{\sqrt{x}(\sqrt{x}+1)}-{1}/{\sqrt{x}+1}]:{\sqrt{x}}/{(\sqrt{x}+1)^2}`
`=[{1}/{\sqrt{x}(\sqrt{x}+1)}-{\sqrt{x}}/{\sqrt{x}(\sqrt{x}+1)}]:{\sqrt{x}}/{(\sqrt{x}+1)^2}`
`={1-\sqrt{x}}/{\sqrt{x}(\sqrt{x}+1)}:{\sqrt{x}}/{(\sqrt{x}+1)^2}`
`={1-\sqrt{x}}/{\sqrt{x}(\sqrt{x}+1)}.{(\sqrt{x}+1)^2}/{\sqrt{x}}`
`={(1-\sqrt{x})(\sqrt{x}+1)}/{x}`
`={1-x}/{x}`
Vậy với `x>0` thì `P={1-x}/{x}`
`b,P={1-x}/{x}={1}/{x}-{x}/{x}={1}/{x}-1`
`P\inZ⇔{1}/{x}\inZ⇔x\in{-1;1}`
Kết hợp ĐKXĐ: `x>0`
`⇒x=1`
Vậy với `x=1` thì `P\inZ`
