Question

va rog mult punctul d de la fiecare dau coroana​

Answer 1

Răspuns:

4. d) $\frac{x+2}{x-1}$ - $\frac{x+1}{x+3}$     se aduce la acelasi numitor

$\frac{( x+2 ) ( x + 3) - ( x+1) ( x-1)}{( x-1) (x+3)}$=  $\frac{(x^{2} + 2x + 3x + 6 ) - (x^{2} -x +x -1) }{(x^{2} -x+ 3x - 3)}$ =

$\frac{ x^{2} + 5x + 6 - x^{2} +1 }{x^{2} +2x -3}$= $\frac{5x +7}{x^{2} +2x -3}$

5. d) $\frac{x-1}{x+1}$ +$\frac{3x-1}{2x+2}$ - $\frac{5x-1 }{3x +3}$= $\frac{x-1}{x+1}$ + $\frac{3x-1}{2(x+1)}$ - $\frac{5x-1 }{3(x +1)}$   aducem la acelasi numitor 6 (x+1)

$\frac{6( x-1) +3(3x-1) - 2(5x-1)}{6(x+1)}$= $\frac{6x- 6 +9x -3 -10x +2}{6(x+1)}$ =$\frac{5x- 7}{6x +6}$

6. d) $\frac{x}{x+y}$ - $\frac{y}{x-y}$    aducem la acelasi numitor (x+y) (x-y)

$\frac{x(x-y) - y(x+y)}{(x+y) (x-y)}$= $\frac{x^{2} -xy -xy -y^{2} }{x^{2} -xy+xy - y^{2} }$= $\frac{x^{2} -2xy- y^{2} }{x^{2} - y^{2} }$

7. d) $\frac{x}{x-3}$ + $\frac{1-x}{3-x}$ - $\frac{x+2}{3-x}$=   aducem la acelasi numitor( x-3) inmultim ultimele doua fractii cu -1

$\frac{x -1 +x +x -2}{x-3}$= $\frac{3x-3 }{x-3}$

Answer 2

Răspuns:

in poza atasata

Sper că te-am ajutat ! :)