Задачи

[aylin96]

[tex](\frac{x+3}{2 })^2-(\frac{x-1}{2 })^2[/tex]

[artur.balabanov]

[tex](\frac{x+3}{2 })^2-(\frac{x-1}{2 })^2[/tex]

[tex](\frac{x+3}{2 })^2-(\frac{x-1}{2 })^2=\frac{(x+3)^2}{4 }-\frac{(x-1)^2}{4 }=\frac{x^2+6x+9-x^2+2x-1}{4 }=\frac{8x+8}{4 }=2x+2[/tex]

[Sara86]

[tex](\frac{x+3}{2 })^2-(\frac{x-1}{2 })^2=\frac{(x+3)^2}{4 }-\frac{(x-1)^2}{4 }=\frac{(x+3)^2-(x-1)^2}{4 }=\frac{(x+3+x-1)(x+3-x+1)}{ 4}=\frac{(2x+2)4}{ 4}=2x+2[/tex]
Нали и така може?
П.С. А и накрая =2(x+1)