Odpowiedź:
[tex]g)\ \ \frac{2}{5}(y+3)^2=\frac{2}{5}(7+3)^2=\frac{2}{5}\cdot10^2=\frac{2}{\not5_{1} }\cdot\not100^2^0=2\cdot20=40\\\\h)\ \ t(t+2)=5(5+2)=5\cdot7=35\\\\\\g)\ \ y(y+3)=-4(-4+3)=-4\cdot(-1)=4\\\\h)\ \ \frac{3}{4}(t-2)^2=\frac{3}{4}(-6-2)^2=\frac{3}{4}\cdot(-8)^2=\frac{3}{\not4_{1}}\cdot\not64^1^6=3\cdot16=48\\\\\\g)\ \ \frac{2}{7}(c-2d)^2=\frac{2}{7}(2-2\cdot(-6))^2=\frac{2}{7}(2+12)^2=\frac{2}{7}\cdot14^2=\frac{2}{\not7_{1}}\cdot\not196^2^8=2\cdot28=56[/tex]
[tex]h)\ \ 0,2d^2(c-d)=0,2\cdot(-5)^2\cdot(0-(-5))=0,2\cdot25\cdot(0+5)=5\cdot5=25[/tex]
