Đáp án+Giải thích các bước giải:
`M=(\frac{1}{\sqrt{x}-1}-\frac{1}{x\sqrt{x}-1}).\frac{3\sqrt{x}-3}{x+\sqrt{x}}(x>0,x\ne1)`
`=(\frac{1}{\sqrt{x}-1}-\frac{1}{(\sqrt{x}-1)(x+\sqrt{x}+1)}).\frac{3(\sqrt{x}-1)}{\sqrt{x}(\sqrt{x}+1)}`
`=(\frac{x+\sqrt{x}+1-1}{(\sqrt{x}-1)(x+\sqrt{x}+1)}).\frac{3(\sqrt{x}-1)}{\sqrt{x}(\sqrt{x}+1)}`
`=\frac{x+\sqrt{x}}{(\sqrt{x}-1)(x+\sqrt{x}+1)}.\frac{3(\sqrt{x}-1)}{\sqrt{x}(\sqrt{x}+1)}`
`=\frac{\sqrt{x}(\sqrt{x}+1).3(\sqrt{x}-1)}{(\sqrt{x}-1)(x+\sqrt{x}+1).\sqrt{x}(\sqrt{x}+1)}`
`=\frac{3}{x+\sqrt{x}+1}`
`b)M\inZZ`
`=>3\vdots x+\sqrt{x}+1`
`=>x+\sqrt{x}+1\inƯ(3)\in{+-1;+-3}`
Mà `x>0`
`=>x+\sqrt{x}+1>1`
`=>x+\sqrt{x}+1=3`
`<=>x+2.\sqrt{x}.1/2 +1/4+3/4=3`
`<=>(x+1/2)^2=9/4`
`<=>(x+1/2)^2=(+-3/2)^2`
`<=>[(x+1/2=3/2),(x+1/2=-3/2):}`
`<=>[(x=1(l)),(x=-2(l)):}`
Vậy không có giá trị `x\inZZ` thỏa mãn `M\inZZ`
