Odpowiedź:
[tex]a)\\\\\frac{2x+3}{5}=\frac{2-x}{3}-x\ \ /\cdot15\\\\\not15^3\cdot\frac{2x+3}{\not5_{1}}=\not15^5\cdot\frac{2-x}{\not3_{1}}-15x\\\\3(2x+3)=5(2-x)-15x\\\\6x+9=10-5x-15x\\\\6x+9=10-20x\\\\6x+20x=10-9\\\\26x=1\ \ /:26\\\\x=\frac{1}{26}[/tex]
[tex]b)\\\\\frac{x+1}{2}-\frac{x-1}{3}=\frac{1}{4}\ \ /\cdot12\\\\\not12^6\cdot\frac{x+1}{\not2_{1}}-\not12^4\cdot\frac{x-1}{\not3_{1}}=\not12^3\cdot\frac{1}{\not4_{1}}\\\\6(x+1)-4(x-1)=3\\\\6x+6-4x+4=3\\\\2x+10=3\\\\2x=3-10\\\\2x=-7\ \ /:2\\\\x=-\frac{7}{2}\\\\x=-3\frac{1}{2}[/tex]
[tex]c)\\\\\frac{4x+7}{2}-\frac{5x-1}{6}=x+5\ \ /\cdot6\\\\\not6^3\cdot\frac{4x+7}{\not2_{1}}-\not6\cdot\frac{5x-1}{\not6}=6x+6\cdot5\\\\3(4x+7)-(5x-1)=6x+30\\\\12x+21-5x+1=6x+30\\\\7x+22=6x+30\\\\7x-6x=30-22\\\\x=8[/tex]
[tex]d)\\\\\frac{x+5}{4}-\frac{2x+1}{8}=\frac{1}{2}(2-x)\\\\\frac{x+5}{4}-\frac{2x+1}{8}=1-\frac{1}{2}x\ \ /\cdot8\\\\\not8^2\cdot\frac{x+5}{\not4_{1}}-\not8\cdot\frac{2x+1}{\not8}=8\cdot1-\not8^4\cdot\frac{1}{\not2_{1}}x\\\\2(x+5)-(2x+1)=8-4x\\\\2x+10-2x-1=8-4x\\\\9=8-4x\\\\4x=8-9\\\\4x=-1\ \ /:4\\\\x=-\frac{1}{4}[/tex]
