Question

ВАРИАНТ 1 +ЗАДАНИЕ СВЕРХУ ПОМОГИТЕ!!!
Даю 20 баллов ​

Answer 1

Ответ на задание сверху , 100% правильное

Answer 2

$\frac{2y^2}{xy+2y^2} -\frac{2xy-x^{2} }{x^{2} -4y^2} =\frac{2y^2}{y(x+2y)} -\frac{2xy-x^{2} }{(x-2y)(x+2y)} =$

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

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

1.

a)      $\frac{3x}{x-4} -\frac{x+8}{x-4}=\frac{3x-x-8}{x-4} =\frac{2x-8}{x-4}=\frac{2(x-4)}{x-4}=2$

б)  $\frac{y-x}{y^2-9} -\frac{x-3}{9-y^2} =\frac{y-x}{y^2-9} -\frac{3-x}{y^2-9} =\frac{y-x+x-3}{(y-3)(y+3)}=\frac{y-3}{(y-3)(y+3)}=\frac{1}{y+3}$

2.

a)    $\frac{b-a}{ab} +\frac{b-a}{b^2} =\frac{b(b-a)+a(b-a)}{ab^2} =\frac{b^2-ab+ab-a^2}{ab^2} =\frac{b^2-a^2}{ab^2}$

б)  $x^{2} +\frac{2x^{2} }{x-2} =\frac{x^{2} (x-2)+2x^{2} }{x-2}= \frac{x^{3} -2x^{2} +2x^{2} }{x-2}= \frac{x^{3} }{x-2}$

3.

$\frac{y^2-2y-3}{y^2-1} +\frac{4}{2y-2} =\frac{y^2-2y-3}{(y-1)(y+1)} +\frac{4}{2(y-1)} =$

$=\frac{2*(y^2-2y-3)+4*(y+1)}{2(y-1)(y+1)} =\frac{2y^2-4y-6+4y+4}{2(y-1)(y+1)}=$

$=\frac{2y^2-2}{2(y-1)(y+1)}=\frac{2(y^2-1)}{2(y-1)(y+1)}=\frac{2(y^2-1)}{2(y^2-1)}=1$ Доказано.