SUBIECTUL 1:
[tex]1)~~m_g = \sqrt{(| \sqrt{5}-3|)(3+\sqrt{5})}= \\
=sqrt{(3-\sqrt{5})(3+\sqrt{5})}=\sqrt{9-5}= \sqrt{4}= \boxed{2}[/tex]
[tex]2) \\
|x+2| \geq 3 =\ \textgreater \ A = (-\infty,~ -5] \bigcup [1, ~+\infty) \\
x \in Z~si~x \geq 3 =\ \textgreater \ B = \{3; ~4; ~5; ~6; ~7;~...\} \\
=\ \textgreater \ A\bigcap B = B ~~deoarece ~~B \subset A [/tex]
[tex]3)~\displaystyle \frac{x+5}{x^2-9} = \frac{-5+5}{5^2-9} =\frac{-5+5}{25-9} =\frac{0}{16} =\boxed{0}[/tex]
4) Rezolvarea o gasesti in primul fisier atasat.
SUBIECTUL 2:
[tex]1a)~~~ x \in R - \{-1 ;~+1\} \\
1b) \\
\displaystyle
E(x) = \frac{x+1}{x^2+1}:\left( \frac{x+3}{4x-4}- \frac{1}{x-1} \right) \cdot \left( 1-\frac{1}{x+1} \right) = \\ \\
= \frac{x+1}{x^2+1}:\left( \frac{x+3}{4(x-1)}- \frac{4}{4(x-1)} \right) \cdot \left( \frac{x+1-1}{x+1} \right) = \\ \\
= \frac{x+1}{x^2+1}: \frac{x+3-4}{4(x-1)}} \cdot \frac{x}{x+1} = \frac{x+1}{x^2+1}: \frac{x-1}{4(x-1)}} \cdot \frac{x}{x+1} =[/tex]
[tex]\displaystyle
=\frac{x+1}{x^2+1}: \frac{1}{4}} \cdot \frac{x}{x+1} = \frac{x+1}{x^2+1}\cdot 4 \cdot \frac{x}{x+1} = \\ \\
=\frac{(x+1) \cdot 4 \cdot x}{(x^2+1)(x+1)}= \boxed{\frac{4x}{x^2+1} }~~~cctd \\ \\
1c)\\
D_{4} = \{ -4;~-2;~-1;~1;~2;~4 \}\\
x^2+1 = -4 ~~=\ \textgreater \ ~ x \notin Z \\
x^2+1 = -2 ~~=\ \textgreater \ ~ x \notin Z \\
x^2+1 = -1 ~~=\ \textgreater \ ~ x \notin Z \\
x^2+1 = 1 ~~~=\ \textgreater \ x = 0 \\
x^2+1 = 2 ~~~=\ \textgreater \ x = 1 \\
x^2+1 = 4 ~~=\ \textgreater \ ~ x \notin Z \\
=\ \textgreater \ x \in \{0;~1\} \\ \\ a[/tex]
2) Rezolvarea o gasesti in ultimele doua fisiere atasate.
