Odpowiedź:
[tex]c)\\\\1\frac{1}{4}x-3(1\frac{1}{3}-2x)=\frac{2-3x}{4}-\frac{x+2}{5}\\\\\frac{5}{4}x-3(\frac{4}{3}-2x)=\frac{2-3x}{4}-\frac{x+2}{5}\\\\\frac{5}{4}x-4+6x=\frac{2-3x}{4}-\frac{x+2}{5}\ \ /*20\\\\25x-80+120x=5(2-3x)-4(x+2)\\\\145x-80=10-15x-4x-8\\\\145x+15x+4x=2+80\\\\164x=82\ \ /:164\\\\x=\frac{82}{164}\\\\x=\frac{1}{2}[/tex]
[tex]d)\\\\\frac{x-1}{2}-\frac{x-2}{4}-\frac{x-4}{8}+1=\frac{5(3-x)}{4}+2x\\\\\frac{x-1}{2}-\frac{x-2}{4}-\frac{x-4}{8}+1=\frac{15-5x}{4}+2x\ \ /*8\\\\4(x-1)-2(x-2)-(x-4)+8=2(15-5x)+16x\\\\4x-4-2x+4-x+4+8=30-10x+16x\\\\x+12=30+6x\\\\x-6x=30-12\\\\-5x=18\ \ /:(-5)\\\\x=-\frac{18}{5}\\\\x=-3\frac{3}{5}[/tex]
