Question

simplificati rapoartele si aflati domeniul lor de definitie
x²+x-2/x²+2x-3 

Answer 1

[tex]\frac{x^2+x-2}{x^2+2x-3}=\\ x^2+2x-3 \neq 0\\ x^2+3x-x-3 \neq 0\\ x(x+3)-(x+3) \neq 0\\ (x+3)(x-1) \neq 0 \Rightarrow x \in R - {-3,1}\\ x^2+x-2=x^2+2x-x-2=x(x+2)-(x+2)=(x+2)(x-1)\\ \frac{x^2+x-2}{x^2+2x-3}=\frac{(x+2)(x-1)}{(x+3)(x-1)} =\frac{x+2}{x+3}[/tex]

Answer 2

x^2+x - 2=x^2-x+2x-2=x(x-1)+2(x-1)=(x-1)(x+2)
x^2+2x-3=x^2+3x-x-3=x(x+3)-(x+3)=(x-1)(x+3)
 => domeniul de definitie ... $x\in{R}-\{+1\;;\;-3\}$
fractia dupa simplificare cu ,,x-1,, devine  ... =(x+2)/(x+3)