`\sqrt{3x+1}+\sqrt{x+1}=8(x>=-1/3)`
`(\sqrt{3x+1}+\sqrt{x+1})^2=8^2`
`3x+1+x+1+2\sqrt{(3x+1)(x+1)}=64`
`4x+2+2\sqrt{(3x+1)(x+1)}=64`
`2x+1+\sqrt{(3x+1)(x+1)}=32`
`\sqrt{(3x+1)(x+1)}=31-2x(x>=31/2)`
`(3x+1)(x+1)=(31-2x)^2`
`3x^2+3x+x+1=4x^2-124x+961`
`x^2-128x+960=0`
`(x^2-8x)+(-120x+960)=0`
`x(x-8)-120(x-8)=0`
`(x-120)(x-8)=0`
`x-120=0` hoặc `x-8=0`
`x=120(L)` hoặc `x=8(tm)`
Vậy: `x=8`
