Odpowiedź:
[tex]\frac{-x}{3}=4-x+\frac{3}{6}\\\frac{-x}{3}=4-x+\frac{1}{2}\\\frac{-x}{3}=\frac{4\cdot \:2}{2}+\frac{1}{2}\\\frac{-x}{3}=\frac{4\cdot \:2+1}{2}\\\frac{-x}{3}=-x+\frac{9}{2}\\\frac{-x}{3}\cdot \:3=-x\cdot \:3+\frac{9}{2}\cdot \:3\\-x=-x\cdot \:3+\frac{27}{2}\\2x=\frac{27}{2}\\\boxed{x=\frac{27}{4}}\\[/tex]
[tex]x-\left(4x-1\right)\left(x+3\right)=13-4x^2\\-4x^2-10x+3=13-4x^2\\-4x^2-10x+3-3=13-4x^2-3\\-4x^2-10x=-4x^2+10\\-4x^2-10x+4x^2=-4x^2+10+4x^2\\-10x=10\\\boxed{x=-1}[/tex]
[tex]x-4x+\frac{1}{3}=\frac{1}{6}+3x\\-3x+\frac{1}{3}=\frac{1}{6}+3x\\-3x+\frac{1}{3}-\frac{1}{3}=\frac{1}{6}+3x-\frac{1}{3}\\-3x=3x-\frac{1}{6}\\-3x-3x=3x-\frac{1}{6}-3x\\-6x=-\frac{1}{6}\\\boxed{x=\frac{1}{36}}[/tex]
