`1, x-3\sqrt{x}=0`
`<=> \sqrt{x}(\sqrt{x}-3)=0`
`<=>`\(\left[ \begin{array}{l}\sqrt{x}=0\\\sqrt{x}-3=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=0\\\sqrt{x}=3\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=0\\x=9\end{array} \right.\)
Vậy `x∈{0;9}`
`2, a+1` $\vdots$ `6`; `b+2007` $\vdots$ `6`
`-> a,b` lẻ `-> a+b` $\vdots$ `2`
Lại có `4^a` $\vdots$ `2` với `∀a`
Nên `4^a+a+b` $\vdots$ `2` (1)
Mặt khác `a+1` $\vdots$ `6`; `b+2007` $\vdots$ `6`
`-> a+1+b+2007` $\vdots$ `6`
`-> (a+b+1)+2007` $\vdots$ `3`
mà `2007` $\vdots$ `3`
`-> a+b+1` $\vdots$ `3`
Ta có : `4^a+a+b=4^a-1+(a+b+1)`
mà `4^a-1` $\vdots$ ` (4-1)` hay `4^a-1` $\vdots$ `3`
`-> 4^a+a+b` $\vdots$ `3` (2)
Từ (1)(2)`-> 4^a+a+b` $\vdots$ `6` (đpcm)
