4.
a)
[tex]\frac{x}{2}=\frac{5}{6} \ \ /\cdot12\\\\6x = 10 \ \ /:6\\\\x = \frac{5}{3}\\\\x = 1\frac{2}{3}[/tex]
b)
[tex]\frac{x}{2} = \frac{x+1}{4} \ \ /\cdot4\\\\2x = x+1\\\\2x-x = 1\\\\x = 1[/tex]
c)
[tex]\frac{2(x-1)}{3}=\frac{x+2}{5} \ \/\cdot15\\\\5\cdot2(x-1)=3(x+2)\\\\10x-10 = 3x+6\\\\10x-3x = 6+10\\\\7x = 16 \ \ /:7\\\\x = \frac{16}{7}\\\\x = 2\frac{2}{7}[/tex]
d)
[tex]1-\frac{2x+5}{3}=4 \ \ /\cdot3\\\\3-(2x+5) = 12\\\\3-2x-5 = 12\\\\-2x-2 = 12\\\\-2x = 12+2\\\\-2x = 14 \ \ /:(-2)\\\\x = -7[/tex]
e)
[tex]\frac{x}{2}-\frac{x}{3} + 1 = \frac{1}{6}(x+6) \ \ /\cdot6\\\\3x-2x+6 = x+6\\\\x-x = 6-6\\\\0 = 0[/tex]
Równanie tożsamościowe, nieskończona ilość rozwiązań.
f)
[tex]\frac{x-1}{4}-\frac{x+1}{2} = -\frac{x}{4} \ \ /\cdot 4\\\\x-1-2(x+1) = -x\\\\x-1-2x-2 = -x\\\\-x+x = 3\\\\0 = 3, \ sprzecznosc\\\\0 \neq 3[/tex]
