Zad.3. Przekształć ilorazy na sumy.
a) (-3 + 6y) : 3 =
[tex]b) (8 {x}^{2} - 4x) \div ( - 2) = [/tex]

[tex] c) \frac{6a + 3b}{3} [/tex]

[tex]d)(5m - n + 1) \div \frac{1}{3} = [/tex]



Odpowiedź :

[tex](-3+6y):3=-3:3+6y:3=-1+2y\\\\(8x^2-4x):(-2)=8x^2:(-2)-4x:(-2)=-4x^2+2x\\\\\frac{6a+3b}{3}=\frac{6a}{3}+\frac{3b}{3}=2a+b\\\\(5m-n+1):\frac{1}{3}=(5m-n+1)*3=5m*3-n*3+1*3=15m-3n+3[/tex]