[tex] \displaystyle\\
3a)\\\\
\frac{x^2-6x+5}{x^2-25}=\frac{x^2-x-5x+5}{x^2-25}=\\\\= \frac{x(x-1)-5(x-1}{(x-5)(x+5)}=\frac{(x-1)(x-5}{(x-5)(x+5)}=\boxed{\frac{x-1}{x+5}}\\\\
3b)\\\\
\frac{x+5-5-1}{x+5}=\frac{x+5-6}{x+5}=\frac{x+5}{x+5}-\frac{6}{x+5}\\\\
D_6 = \{-6;-3;-2;-1;1;2;3;6\}\\\\
x+5 = -6 \to x = -11\\
x+5=-3 \to x=-8\\
x+5=-2\to x=-7\\
x+5=-1 \to x=-6\\
x+5=1\to x=-4\\
x+5=2\to x=-3\\
x+5=3\to x=-2\\
x+5=6\to x=1\\
\boxed{M=\{-11;-8;-7;-6;-4;-3;-2;1\}} [/tex]
[tex] \displaystyle\\
4)\\
\sqrt{x^2-2\sqrt{2}x+2} +\sqrt{y^2+\sqrt{12}y+3} =0\\\\
\sqrt{x^2-2\sqrt{2}x+2} +\sqrt{y^2+2\sqrt{3}y+3} =0\\\\
\sqrt{\Big(x-\sqrt{2}\Big)^2} +\sqrt{\Big(y+\sqrt{3}\Big)^2} =0\\\\
\text{Avem 2 patrate care sunt mai mari sau egale cu 0}\\
\text{Pentru ca suma sa fie egala cu zero trebuie ca fiecare patrat sa fie 0.}\\
x-\sqrt{2}=0 \to \boxed{x = \sqrt{2}}\\
y+\sqrt{3}=0 \to \boxed{y= -\sqrt{3}} [/tex]
[tex] \displaystyle\\
5)\\
\frac{x+0}{1}+\frac{x+1}{2}+\frac{x+2}{3}+...+\frac{x+2016}{2017}=2017\\\\
\text{Observam ca:}\\
\text{Avem 2017 fractii}\\
\text{Numitorii sunt numere consecutive.}\\
\text{Numaratorii sunt numere consecutive}\\
\text{Cautam o valoare pentru x a. i. fractiile sa fie echivalente.}\\\\
\frac{x+1}{2}=\frac{x+2}{3}\\
3(x+1)=2(x+2)\\
3x+3=2x+4\\
3x-2x=4-3\\
\boxed{x=1}\\
\text{Verificam:}\\
\frac{1+0}{1}=1\\
\frac{1+1}{2}=1\\
\frac{1+2}{3}=1\\
\frac{1+2016}{2017}=1\\
1+1+1+...+1=2017
[/tex]
