Question

Ma poate ajuta cineva??? Va rog.

Answer 1

$\displaystyle\\ 3a)\\\\ \frac{x^2-6x+5}{x^2-25}=\frac{x^2-x-5x+5}{x^2-25}=\\\\= \frac{x(x-1)-5(x-1}{(x-5)(x+5)}=\frac{(x-1)(x-5}{(x-5)(x+5)}=\boxed{\frac{x-1}{x+5}}\\\\ 3b)\\\\ \frac{x+5-5-1}{x+5}=\frac{x+5-6}{x+5}=\frac{x+5}{x+5}-\frac{6}{x+5}\\\\ D_6 = \{-6;-3;-2;-1;1;2;3;6\}\\\\ x+5 = -6 \to x = -11\\ x+5=-3 \to x=-8\\ x+5=-2\to x=-7\\ x+5=-1 \to x=-6\\ x+5=1\to x=-4\\ x+5=2\to x=-3\\ x+5=3\to x=-2\\ x+5=6\to x=1\\ \boxed{M=\{-11;-8;-7;-6;-4;-3;-2;1\}}$

$\displaystyle\\ 4)\\ \sqrt{x^2-2\sqrt{2}x+2} +\sqrt{y^2+\sqrt{12}y+3} =0\\\\ \sqrt{x^2-2\sqrt{2}x+2} +\sqrt{y^2+2\sqrt{3}y+3} =0\\\\ \sqrt{\Big(x-\sqrt{2}\Big)^2} +\sqrt{\Big(y+\sqrt{3}\Big)^2} =0\\\\ \text{Avem 2 patrate care sunt mai mari sau egale cu 0}\\ \text{Pentru ca suma sa fie egala cu zero trebuie ca fiecare patrat sa fie 0.}\\ x-\sqrt{2}=0 \to \boxed{x = \sqrt{2}}\\ y+\sqrt{3}=0 \to \boxed{y= -\sqrt{3}}$

$\displaystyle\\ 5)\\ \frac{x+0}{1}+\frac{x+1}{2}+\frac{x+2}{3}+...+\frac{x+2016}{2017}=2017\\\\ \text{Observam ca:}\\ \text{Avem 2017 fractii}\\ \text{Numitorii sunt numere consecutive.}\\ \text{Numaratorii sunt numere consecutive}\\ \text{Cautam o valoare pentru x a. i. fractiile sa fie echivalente.}\\\\ \frac{x+1}{2}=\frac{x+2}{3}\\ 3(x+1)=2(x+2)\\ 3x+3=2x+4\\ 3x-2x=4-3\\ \boxed{x=1}\\ \text{Verificam:}\\ \frac{1+0}{1}=1\\ \frac{1+1}{2}=1\\ \frac{1+2}{3}=1\\ \frac{1+2016}{2017}=1\\ 1+1+1+...+1=2017$