Question

Dau 100 de puncte pentru 38 si 39

Answer 1

[tex]E(x)=\frac{x^3+2x^2-3x-6}{x^2-4}, x \in R-\{-2,2\}\\ a)E(x)=\frac{x^2(x+2)-3(x+2)}{(x-2)(x+2)}\\ E(x)=\frac{(x+2)(x^2-3)}{(x-2)(x+2)}=\frac{x^2-3}{x-2}\\ b)E(-1)\cdotE(0)\cdotE(1)=\frac{(-1)^2-3}{-1-2}\cdot \frac{0-3}{0-2}\cdot\frac{1-3}{1-2}=\\ =\frac{-2}{-3}\cdot\frac{-3}{-2}\cdot\frac{-2}{-1}=\boxed{\boxed{2}}\\ c)(n-2)\cdot E(n)+6(n+2)=pp,n \geq 3\\ (n-2)\cdot \frac{n^2-3}{n-2}+6n+12=\\ =n^2-3+6n+12=\\ =n^2+6n+9=\boxed{\boxed{(n+3)^2}}\\ 39.F(x)=\frac{x^2+6x+9}{x^2+x-6}\\ a)*x^2+6x+9=(x+3)^2\\[/tex][tex]x^2+x-6=x^2+3x-2x-6=x(x+3)-2(x+3)=\\ =(x+3)(x-2)\\ F(x)=\frac{(x+3)^2}{(x+3)(x-2)}\\ F(x)=\frac{x+3}{x-2}\\ F(x)=\frac{x-2+5}{x-2}\\ F(x)=\frac{x-2}{x-2}+\frac{5}{x-2}\\ F(x)=1+\frac{5}{x-2} \vee x \in R-\{-3,2\}\\ b)F(x)\in N\\ 1+\frac{5}{x-2}\in N\\ Dar\ stim\ ca\ 1\ \in N,deci\ trebuie\ doar\ pentru\ \frac{5}{x-2}\\ Vom\ avea\ :\frac{5}{x-2} \in N\Rightarrow x-2 \in D_5\\ x-2 \in \{1,5\} \\ x \in \{3,7\}\\ c)S=F(1\cdot2+2)+F(2\cdot3+2)+F(3\cdot4+2)+...+F(2012\cdot2013+2\\ [/tex][tex]S=1+\frac{5}{1\cdot2+2-2}+1+\frac{5}{2\cdot3+2-2}+...+1+\frac{5}{2012\cdot2013+2-2}\\ S=1+\frac{5}{1\cdot2}+1+\frac{5}{2\cdot3}+...+1+\frac{5}{2012\cdot2013}\\ S=2012+5(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{2012\cdot2013})\\ S=2012+5(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{2012}-\frac{1}{2013})\\ S=2012+5(1-\frac{1}{2013})\\ S=2012+5\cdot\frac{2012}{2013}\\ S=2012(1+\frac{5}{2013})\\ S=2012\cdot \frac{2018}{2013}\\ [/tex]
[tex]S=\frac{4051056+2018}{2013}\\ \boxed{\boxed{S=\frac{4052174}{2013}}}[/tex]