Răspuns:
[tex]\boldsymbol{ a) \ \red{\dfrac{x-2}{x+2}}, \ b) \ \red{\dfrac{7}{3}}, \ c) \ \red{\dfrac{x-2}{x-3}}}[/tex]
Explicație pas cu pas:
[tex]a) \ \dfrac{x^2-4}{(x+2)^2} = \dfrac{(x-2)(x+2)}{(x+2)^2}^{(x+2} = \dfrac{x-2}{x+2}, \ \ \ \ x \neq -2[/tex]
❇
[tex]b) \ \dfrac{7x+14}{3(x+2)} = \dfrac{7(x+2)}{3(x+2)}^{(x+2} = \dfrac{7}{3}, \ \ \ \ x \neq -2[/tex]
❇
[tex]c) \ \dfrac{x^2-3x+2}{x^2-4x+3} = \dfrac{(x-1)(x-2)}{(x-1)(x-3)}^{(x-1} = \dfrac{x-2}{x-3}, \ \ \ \ x \neq \{1,3\}[/tex]
unde
x²-3x+2 = x²-x-2x+2 = x(x-1)-2(x-1) = (x-1)(x-2)
x²-4x+3 = x²-x-3x+3 = x(x-1)-3(x-1) = (x-1)(x-3)
