-2;-1;0;1

\(\sqrt{5x}-2\le4\Rightarrow\sqrt{5x}\le6.\) 

I5xI<=36

\(\orbr{\begin{cases}x< =\frac{36}{5}\approx7^+\\x>=\frac{-36}{5}\approx7^-\end{cases}}\),

S={-7,-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,7)

\(\sqrt{5x}-2< =4\)                                       ĐK:\(\sqrt{5x}>0\)<=> 5x > 0 <=> x>0

<=>\(\sqrt{5x}< =4+2\)

<=>\(\sqrt{5x}\)<= 6

<=> 5x <= \(6^2\)

<=>5x <= 36

<=> x <= \(\frac{36}{5}\)

<=> x <= 7,2

\(\sqrt{5x-2}\le4\)

<=>\(\begin{cases}5x-2\ge0\\5x-2\le16\end{cases}\)<=> \(\begin{cases}x\ge\frac{2}{5}\\x\le\frac{18}{5}\end{cases}\)

<=>x=1,2,3

\(x^2+4x-5=0\)

\(\Leftrightarrow x^2+5x-x-5=0\)

\(\Leftrightarrow x^2+5x-\left(x+5\right)=0\)

\(\Leftrightarrow x\left(x+5\right)-\left(x+5\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+5\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-1=0\\x+5=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=-5\end{cases}}}\)

để n/n+6 là số nguyên thi N chia het cho n+6 

mà [n+6] chia hết cho[ơn+6]

=>[n+6]-n chia hết cho [n+6]

=>6 chia hết cho n+6 =>n+6 = 2 3 6 -2 -3 -6

=>n=-4 -3 0 -8 -9 -12

ooop banana

ta có : \(\frac{n}{n+3}=\frac{\left(n+3\right)-3}{n+3}\)

vì \(\left(n+3\right)⋮\left(n+3\right)\)để \(\frac{\left(n+3\right)-3}{n+3}\)nguyên \(\Leftrightarrow-3⋮\left(n+3\right)\Leftrightarrow\left(n+3\right)\inƯ\left(-3\right)\RightarrowƯ\left(-3\right)=1;-1;3;-3\)

\(\Rightarrow n+3=1\Rightarrow n=-2\)

\(\Rightarrow n+3=-1\Rightarrow n=-4\)

\(\Rightarrow n+3=3\Rightarrow n=0\)

\(\Rightarrow n+3=-3\Rightarrow n=-6\)

vậy \(S=-6;-4;-2;0\)

hiểu chết liền

Ta có: 
n/(n + 3) = (n + 3 - 3)/(n + 3) = 1 - 3/(n + 3)
Để n/(n + 3) là số nguyên thì 3 ⋮ (n + 3)
Suy ra: (n + 3) ∈ Ư(3)
Ta có: Ư(3) = {-3; -1; 1; 3}
Suy ra: n = {-6; -4; -2; 0}
Vậy: S = {-6;-4;-2;0}