Răspuns:
4. d) [tex]\frac{x+2}{x-1}[/tex] - [tex]\frac{x+1}{x+3}[/tex] se aduce la acelasi numitor
[tex]\frac{( x+2 ) ( x + 3) - ( x+1) ( x-1)}{( x-1) (x+3)}[/tex]= [tex]\frac{(x^{2} + 2x + 3x + 6 ) - (x^{2} -x +x -1) }{(x^{2} -x+ 3x - 3)}[/tex] =
[tex]\frac{ x^{2} + 5x + 6 - x^{2} +1 }{x^{2} +2x -3}[/tex]= [tex]\frac{5x +7}{x^{2} +2x -3}[/tex]
5. d) [tex]\frac{x-1}{x+1}[/tex] +[tex]\frac{3x-1}{2x+2}[/tex] - [tex]\frac{5x-1 }{3x +3}[/tex]= [tex]\frac{x-1}{x+1}[/tex] + [tex]\frac{3x-1}{2(x+1)}[/tex] - [tex]\frac{5x-1 }{3(x +1)}[/tex] aducem la acelasi numitor 6 (x+1)
[tex]\frac{6( x-1) +3(3x-1) - 2(5x-1)}{6(x+1)}[/tex]= [tex]\frac{6x- 6 +9x -3 -10x +2}{6(x+1)}[/tex] =[tex]\frac{5x- 7}{6x +6}[/tex]
6. d) [tex]\frac{x}{x+y}[/tex] - [tex]\frac{y}{x-y}[/tex] aducem la acelasi numitor (x+y) (x-y)
[tex]\frac{x(x-y) - y(x+y)}{(x+y) (x-y)}[/tex]= [tex]\frac{x^{2} -xy -xy -y^{2} }{x^{2} -xy+xy - y^{2} }[/tex]= [tex]\frac{x^{2} -2xy- y^{2} }{x^{2} - y^{2} }[/tex]
7. d) [tex]\frac{x}{x-3}[/tex] + [tex]\frac{1-x}{3-x}[/tex] - [tex]\frac{x+2}{3-x}[/tex]= aducem la acelasi numitor( x-3) inmultim ultimele doua fractii cu -1
[tex]\frac{x -1 +x +x -2}{x-3}[/tex]= [tex]\frac{3x-3 }{x-3}[/tex]
