Zadanie 1
[tex]2a+4a=6a\\\\3b\cdot4b=12b^2\\\\\frac{8c}{4c}=2\\\\\frac{8c}{2}=4c\\\\\frac{8}{4c}=\frac{2}{c}[/tex]
Zadanie 2
[tex]a(1+a)=a\cdot1+a\cdot a=a^2+a\\\\1,5(4a-1)=1,5\cdot4a-1,5\cdot1=6a-1,5\\\\(\frac{1}{2}-c)c=\frac{1}{2}\cdot c-c\cdot c=-c^2+\frac{1}{2}c\\\\(1-d)(1+d)=1+d-d-d^2=-d^2+1[/tex]
Zadanie 3
[tex]\frac{2a+4c}{2}=\frac{2(a+2c)}{2}=a+2c\\\\\frac{24x-4y}{8}=\frac{8(3x-0,5y)}{8}=3x-0,5y\\\\\frac{36c+4c^2}{2}=\frac{2(18c+2c^2)}{2}=18c+2c^2\\\\\frac{25d^3-5d^2}{20}=\frac{20(1,25d^3-0,25d^2)}{20}=1,25d^3-0,25d^2[/tex]
