`***`Lời giải`***`
`M=\frac{\sqrt{x}+1}{x+\sqrt{x}+1}`
ĐKXĐ: `x≥0`
Ta có: `M=3/7`
`=>\frac{\sqrt{x}+1}{x+\sqrt{x}+1}=3/7`
`<=>\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-3/7=0`
`<=>\frac{7(\sqrt{x}+1)-3(x+\sqrt{x}+1)}{7(x+\sqrt{x}+1)}=0`
`=>7(\sqrt{x}+1)-3(x+\sqrt{x}+1)=0`
`<=>7\sqrt{x}+7-3x-3\sqrt{x}-3=0`
`<=>-3x+4\sqrt{x}+4=0`
`<=>3x-4\sqrt{x}-4=0`
`<=>3x-6\sqrt{x}+2\sqrt{x}-4=0`
`<=>3\sqrt{x}(\sqrt{x}-2)+2(\sqrt{x}-2)=0`
`<=>(3\sqrt{x}+2)(\sqrt{x}-2)=0`
`+)3\sqrt{x}+2=0<=>3\sqrt{x}=-2<=>\sqrt{x}=-2/3`(vô nghiệm)
`+)\sqrt{x}-2=0<=>\sqrt{x}=2<=>x=4(N)`
Vậy` x=4 `để `M=3/7`
