Szczegółowe wyjaśnienie:
[tex]a)\ 2x-(x+3)=6+2x\\\\2x-x-3=6+2x\\\\x-3=6+2x\qquad|+3-2x\\\\-x=9\\\\\huge\boxed{x=-9}[/tex]
[tex]b)\ \dfrac{1}{2}(x+5)=\dfrac{2}{3}x+1\qquad|\cdot6\\\\6\!\!\!\!\diagup^3\cdot\dfrac{1}{2\!\!\!\!\diagup_1}(x+5)=6\!\!\!\!\diagup^2\dfrac{2}{3\!\!\!\!\diagup_1}x+6\cdot1\\\\3(x+5)=2\cdot2x+6\\\\3x+15=4x+6\qquad|-15-4x\\\\-x=-9\\\\\huge\boxed{x=9}[/tex]
[tex]c)\ \dfrac{x+1}{2}-\dfrac{x}{5}=\dfrac{5x+2}{10}\qquad|\cdot10\\\\10\!\!\!\!\!\diagup^5\cdot\dfrac{x+1}{2\!\!\!\!\diagup_1}-10\!\!\!\!\!\diagup^2\cdot\dfrac{x}{5\!\!\!\!\diagup_1}=10\!\!\!\!\!\diagup^1\cdot\dfrac{5x+2}{10\!\!\!\!\!\diagup_1}\\\\5(x+1)-2x=5x+2\\\\5x+5-2x=5x+2\\\\3x+5=5x+2\qquad|-5-5x\\\\-2x=-3\qquad|:(-2)\\\\\huge\boxed{x=\dfrac{3}{2}}[/tex]
[tex]d)\ x(1+x)=-x^2+2(3x+x^2)\\\\x+x^2=-x^2+6x+2x^2\\\\x+x^2=x^2+6x\qquad|-x^2\\\\x=6x\qquad|-6x\\\\-5x=0\qquad|:(-5)\\\\\huge\boxed{x=0}[/tex]
