Question

Помогите пж............

Answer 1

Объяснение:

1) а)$\frac{x^2-2xy+4y^2}{x-2y}+\frac{x^2+2xy+y^2}{x+2y}=\frac{(x+2y)(x^2-2xy+4y^2)+(x-2y)(x^2+2xy+y^2)}{(x-2y)(x+2y)}=\\\frac{x^3+8y^3+x^3+2x^2y+xy^2-2x^2y-4xy^2-2y^3}{x^2-4y^2}=\frac{2x^3+6y^3-3xy^2}{x^2-4xy^2}$

б)$\frac{x^2-2xy+4y^2}{x-2y}-\frac{x^2+2xy+4y^2}{x+2y}=\frac{(x+2y)(x^2-2xy+4y^2)-(x-2y)(x^2+2xy+4y^2)}{(x-2y)(x+2y)}=\frac{x^3+8y^3-(x^3-8y^3)}{x^2-4y^2}=\frac{x^3+8y^3-x^3+8y^3}{x^2-4y^2}=\frac{16y^3}{x^2-4y^2}$2) а)$\frac{a-b}{a+b}-\frac{a^2+b^2}{a^2-b^2}+\frac{a+b}{a-b}=\frac{a-b}{a+b}-\frac{a^2+b^2}{(a-b)(a+b)}+\frac{a+b}{a-b}=\frac{(a-b)^2-(a^2+b^2)+(a+b)^2}{(a-b)(a+b)}=\frac{a^2-2ab+b^2-a^2-b^2+a^2+2ab+b^2}{a^2-b^2}=\frac{a^2+b^2}{a^2-b^2}$б)$\frac{1}{2-y}-\frac{1}{2+y}-\frac{y}{4-y^2} +\frac{y^2+4}{2y^3-8y}=\frac{1}{-(y-2)}-\frac{1}{2+y}-\frac{y}{(2-y)(2+y)}+\frac{y^2+4}{2y(y^2-4)}=-\frac{1}{y-2}-\frac{1}{2+y}-\frac{y}{-(y-2)(2+y)}+\frac{y^2+4}{2y(y-2)(y+2)}=-\frac{1}{y-2}-\frac{1}{2+y}+\frac{y}{(y-2)(2+y)}+\frac{y^2+4}{2y(y-2)(2+y)}=\frac{-2y(2+y)-2y(y-2)+2y^2+y^2+4}{2y(y-2)(2+y)}=$$\frac{-4y-2y^2-2y^2+4y+2y^2+y^2+4}{2y(y-2)(2+y)}=\frac{-y^2+4}{2y(y-2)(2+y)}=\frac{4-y^2}{2y(y-2)(2+y)}=\frac{(2-y)(2+y)}{2y(y-2)(2+y)}=\frac{-(y-2)*1}{2y(y-2)}=\frac{-1}{2y}=-\frac{1}{2y}$