Explicație pas cu pas:
a)
intrucat x ∈ Z, iar sqrt(2) ≈ 1.41
[tex]x^{2} \leq 2 <=> |x| \leq \sqrt{2} <=> -\sqrt{2} \leq x\leq \sqrt{2} => A = {-1, 0, 1}[/tex]
b)
(x+1)∈D₈={1, 2, 4, 8} ( se scade 1) => x∈D₈={0, 1, 3, 7} => A = {0, 1, 3, 7}
c)
x²- 1 = -3 sau x² - 1 = 3 ( conform definitie modulului) => [tex]\left \{ {{x^{2}= -2} \atop {x^{2}= 4}} \right.[/tex]
x ∈ R deci x²≠-2 => A = { [tex]-\sqrt{2} , \sqrt{2}[/tex]}
d)
[tex]4^{x} = 2^{2x}[/tex]
notam [tex]2^{x} = t[/tex]
ecuatia devine: [tex]t^{2} - 7t + 12 = 0 => (t-4)(t-3) = 0 => \left \{ {{t=4} \atop {t=3}} \right.[/tex]
dar [tex]2^{x} = t[/tex] => [tex]\left \{ {{x=2} \atop {x=log2(3)}} \right.[/tex] => A= {2, log2(3)}
