Odpowiedź:
[tex]a)\ \ \dfrac{\not10^2(3x+2)}{\not5_{1}}+\dfrac{6x-18}{3}=2(3x+2)+\dfrac{\not3(2x-6)}{\not3}=6x+4+(2x-6)=\\\\\\=6x+4+2x-6=8x-2\\\\\\b)\ \ 2\cdot\dfrac{\not15^5(2x-1)}{\not3_{1}}-\dfrac{8-12x}{4}=2\cdot5(2x-1)-\dfrac{\not4(2-3x)}{\not4}=10(2x-1)-(2-3x)=\\\\\\=20x-10-2+3x=23x-12\\\\\\c)\ \ 2\cdot\dfrac{3x-6}{3}-3\cdot\dfrac{6-18x}{6}=2\cdot\dfrac{\not3(x-2)}{\not3}-3\cdot\dfrac{\not6(1-3x)}{\not6}=2(x-2)-3(1-3x)=\\\\\\=2x-4-3+9x=11x-7[/tex]
[tex]d)\ \ \dfrac{12x-4}{4}+2\cdot\dfrac{15x-10}{5}=\dfrac{\not4(3x-1)}{\not4}+2\cdot\dfrac{\not5(3x-2)}{\not5}=3x-1+2(3x-2)=\\\\\\=3x-1+6x-4=9x-5[/tex]
