Đáp án:
` a\in {0;7+4\sqrt{3};7-4\sqrt{3}}`
Giải thích các bước giải:
`N={6\sqrt{a}}/{(\sqrt{a}+1)^2}` `(a\ge 0)`
Với mọi `a\ge 0`
`=>`$\begin{cases}\sqrt{a}\ge 0\\(\sqrt{a}+1)^2\ge 1> 0\end{cases}$
`=>N={6\sqrt{a}}/{(\sqrt{a}+1)^2}\ge 0`
$\\$
`N={12\sqrt{a}}/{2(\sqrt{a}+1)^2}`
`N={12\sqrt{a}-3(\sqrt{a}+1)^2+3(\sqrt{a}+1)^2}/{2(\sqrt{a}+1)^2}`
`N={12\sqrt{a}-3(a+2\sqrt{a}+1)}/{2(\sqrt{a}+1)^2}+{3(\sqrt{a}+1)^2}/{2(\sqrt{a}+1)^2}`
`N={-3(a-2\sqrt{a}+1)}/{2(\sqrt{a}+1)^2}+3/2`
`N={-3(\sqrt{a}-1)^2}/{2(\sqrt{a}+1)^2}+3/2`
Với mọi `x\ge 0`
`=>(\sqrt{a}-1)^2\ge 0; (\sqrt{a}+1)^2\ge 1>0`
`=>-3(\sqrt{a}-1)^2\le 0`
`=>N={-3(\sqrt{a}-1)^2}/{2(\sqrt{a}+1)^2}+3/2\le 3/2`
`=>0\le N\le 3/2`
Để `N\in ZZ=>N\in {0;1}`
+) `TH: N=0`
`<=>{6\sqrt{a}}/{(\sqrt{a}+1)^2}=0`
`<=>6\sqrt{a}=0`
`<=>\sqrt{a}=0<=>a=0\ (thỏa\ đk)`
+) `TH: N=1`
`<=>{6\sqrt{a}}/{(\sqrt{a}+1)^2}=1`
`<=>6\sqrt{a}=(\sqrt{a}+1)^2`
`<=>6\sqrt{a}=a+2\sqrt{a}+1`
`<=>a-4\sqrt{a}+1=0`
`<=>a-4\sqrt{a}+4-3=0`
`<=>(\sqrt{a}-2)^2=3`
`<=>`$\left[\begin{array}{l}\sqrt{a}-2=\sqrt{3}\\\sqrt{a}-2=-\sqrt{3}\end{array}\right.$
`<=>`$\left[\begin{array}{l}\sqrt{a}=2+\sqrt{3}\\\sqrt{a}=2-\sqrt{3}\end{array}\right.$
`<=>`$\left[\begin{array}{l}a=(2+\sqrt{3})^2=4+4\sqrt{3}+3\\a=(2-\sqrt{3})^2=4-4\sqrt{3}+3\end{array}\right.$
`<=>`$\left[\begin{array}{l}a=7+4\sqrt{3}\\a=7-4\sqrt{3}\end{array}\right.$
Vậy `a\in {0;7+4\sqrt{3};7-4\sqrt{3}}` thì `N\in ZZ`
