1. a) Solve in $Z$ the equation: $5 \cdot(2 \cdot|3 x-4|+4)-30=10$
Solution: $5(2|3 x-4|+4)=40 \rightarrow 2|3 x-4|+4=8 \rightarrow|3 x-4|=2 \rightarrow x=2 \in \mathbb{Z}$ and $x=\frac{2}{3} \notin \mathbb{Z}$ (2 points)

b) If $x=\sqrt{2010+(2+4+6+\cdots+4018)}$, show that $x \in \mathbf{N}$.

Solution: $x=\sqrt{2010+2(1+2+\cdots 2009)}=\sqrt{2010+2009 \cdot 2010}=\sqrt{2010 \cdot 2010}=$

$2010 \in \mathbb{N}$

(3 points)

c) Let a, b, c, d be positive real numbers such that abcd=1. Calculate:

$\mathrm{E}=\frac{7+a}{1+a+a b+a b c}+\frac{7+b}{1+b+b c+b c d}+\frac{7+c}{1+c+c d+c d a}+\frac{7+d}{1+d+d a+d a b}$.

Solution:

d)

$\frac{7+a}{1+a+a b+a b c}+\frac{7+b}{1+b+b c+b c d}+\frac{7+c}{1+c+c d+c d a}+\frac{7+d}{1+d+d a+d a b}={ }^{b c)}$
$\frac{7 b+b c+7+b}{1+b+b c+b c d}=\frac{8+8 b+8 b c+8 b c d}{1+b+b c+b c d}=8$.