Question

7x supra 2x+2 -3x+5 supra 3x+3 +2-3x supra x+1- x+6 supra 6x+6

Exercitiul 5 punctul b

Answer 1

$\frac{7x}{2x+2}- \frac{3x+5}{3x+3}+ \frac{2-3x}{x-1}- \frac{x+6}{6x+6}= [tex] \frac{7x}{2(x+1)}- \frac{3x+5}{3x+1}+ \frac{2-3x}{x-1}- \frac{x+6}{6(x+1)}$= [/tex]$\frac{3*7x}{3*2(x+1)}- \frac{2(3x+5)}{2*3(x+1)}+ \frac{6(2-3x)}{6(x+1)}- \frac{x+6}{6(x+1)}=$$\frac{21x}{6(x+1)}- \frac{2(3x+5)}{6(x+1)}+ \frac{6(2-3x)}{6(x+1)}- \frac{x+6}{6(x+1)}=$$\frac{21x-2(3x+5)+6(2-3x)-(x+6)}{6(x+1)}= \frac{21x-6x-10+12-18x-x-6}{6(x+1)}=$$\frac{-4x-4}{6(x+1)}= \frac{-4(x-(-1))}{6(x+1)}= \frac{-2(x+1)}{3(x+1)}= \frac{-2}{3}=- \frac{2}{3}$