`ĐKXĐ:x>0,x\ne1`
`a,M={x\sqrt{x}-1}/{x-\sqrt{x}}-{x\sqrt{x}+1}/{x+\sqrt{x}}+(\sqrt{x}-{1}/{\sqrt{x}})({\sqrt{x}+1}/{\sqrt{x}-1}+{\sqrt{x}-1}/{\sqrt{x}+1})`
`={(\sqrt{x}-1)(x+\sqrt{x}+1)}/{\sqrt{x}(\sqrt{x}-1)}-{(\sqrt{x}+1)(x-\sqrt{x}+1)}/{\sqrt{x}(\sqrt{x}+1)}+({x}/{\sqrt{x}}-{1}/{\sqrt{x}})[{(\sqrt{x}+1)(\sqrt{x}+1)}/{(\sqrt{x}+1)(\sqrt{x}-1)}+{(\sqrt{x}-1)(\sqrt{x}-1)}/{(\sqrt{x}+1)(\sqrt{x}-1)}]`
`={x+\sqrt{x}+1}/{\sqrt{x}}-{x-\sqrt{x}+1}/{\sqrt{x}}+{x-1}/{\sqrt{x}}.{(\sqrt{x}+1)(\sqrt{x}+1)+(\sqrt{x}-1)(\sqrt{x}-1)}/{(\sqrt{x}+1)(\sqrt{x}-1)}`
`={x+\sqrt{x}+1-(x-\sqrt{x}+1)}/{\sqrt{x}}+{(\sqrt{x}+1)(\sqrt{x}-1)}/{\sqrt{x}}.{(\sqrt{x}+1)^2+(\sqrt{x}-1)^2}/{(\sqrt{x}+1)(\sqrt{x}-1)}`
`={x+\sqrt{x}+1-x+\sqrt{x}-1}/{\sqrt{x}}+{(\sqrt{x}+1)^2+(\sqrt{x}-1)^2}/{\sqrt{x}}`
`={2\sqrt{x}}/{\sqrt{x}}+{x+2\sqrt{x}+1+x-2\sqrt{x}+1}/{\sqrt{x}}`
`={2\sqrt{x}}/{\sqrt{x}}+{2x+2}/{\sqrt{x}}`
`={2x+2\sqrt{x}+2}/{\sqrt{x}}`
Vậy với `x>0,x\ne1` thì `M={2x+2\sqrt{x}+2}/{\sqrt{x}}`
`b,M=6`
`⇔{2x+2\sqrt{x}+2}/{\sqrt{x}}=6`
`⇔{2x+2\sqrt{x}+2}/{\sqrt{x}}={6\sqrt{x}}/{\sqrt{x}}`
`⇒2x+2\sqrt{x}+2=6\sqrt{x}`
`⇔2x-4\sqrt{x}+2=0`
`⇔2(x-2\sqrt{x}+1)=0`
`⇔x-2\sqrt{x}+1=0`
`⇔(\sqrt{x}-1)^2=0`
`⇔\sqrt{x}-1=0`
`⇔\sqrt{x}=1`
`⇔x=1(KTM)`
Vậy không tìm được giá trị `x` thỏa mãn `M=6`
