Răspuns:
[tex]\boldsymbol{a) \ \red{ -1 }} ; \ \boldsymbol{b) \ \red{ 2 }} ; \ \boldsymbol{c) \ \red{ -2 }}[/tex]
Explicație pas cu pas:
[tex]a) \ \dfrac{3}{x-3}+\dfrac{x}{3-x} = \dfrac{3}{x-3}-\dfrac{x}{x-3} = \dfrac{3-x}{x-3} =\\[/tex]
[tex]= - \dfrac{x-3}{x-3} = -1, \ \ x\neq3[/tex]
[tex]b) \ \dfrac{5x+1}{x-2}+\dfrac{3x+5}{2-x} = \dfrac{5x+1}{x-2}-\dfrac{3x+5}{x-2} = \dfrac{5x+1-3x-5}{x-2} =\\[/tex]
[tex]= \dfrac{2x-4}{x-2} = \dfrac{2(x-2)}{x-2} = 2, \ \ x\neq2[/tex]
[tex]c) \ \dfrac{x^2-2x+3}{2x-1}+\dfrac{x^2+2x+1}{1-2x} = \dfrac{x^2-2x+3}{2x-1}-\dfrac{x^2+2x+1}{2x-1} =\\[/tex]
[tex]= \dfrac{x^2-2x+3-x^2-2x-1}{2x-1} = \dfrac{-4x+2}{2x-1} = -\dfrac{4x-2}{2x-1}\\[/tex]
[tex]= -\dfrac{2(2x-1)}{2x-1} = - 2, \ \ x\neq \dfrac{1}{2}[/tex]
⇒ sumele sunt constante
