Виконайте дію:
1) $\frac{2a^2-2a}{4a^2-b^2}+\frac{b(a-1)}{b^2-4a^2};$
2) $\frac{x+3}{xy-x^2}+\frac{y+3}{xy-y^2};$
3) $\frac{x^2+x-3}{x^2-9}-\frac{3}{2x-6}-1;$
4) $\frac{ax-2x}{3y+6}:\frac{4-a^2}{y^2+4y+4}.$
Розв'язок:
1) $\frac{2a^2-2a}{4a^2-b^2}+\frac{b\left(a-1\right)}{b^2-4a^2}=$
$= \frac{2a\left(a-1\right)}{4a^2-b^2}-\frac{b\left(a-1\right)}{4a^2-b^2}=$
$= \frac{\left(a-1\right)\left(2a-b\right)}{\left(2a-b\right)\left(2a+b\right)}=\frac{a-1}{2a+b};$
2) $\frac{x+3}{xy-x^2}+\frac{y+3}{xy-y^2}=$
$= \frac{x+3}{x\left(y-x\right)}+\frac{y+3}{y\left(x-y\right)}=$
$= \frac{x+3}{x\left(y-x\right)}-\frac{y+3}{y\left(y-x\right)}=$
$= \frac{y(x+3)-x(y+3)}{xy(y-x)}=\frac{xy+3y-xy-3x}{xy(y-x)}=$
$= \frac{3(y-x)}{xy(y-x)}=\frac{3}{xy};$
3) $\frac{x^2+x-3}{x^2-9}-\frac{3}{2x-6}-1=$
$= \frac{x^2+x-3}{\left(x-3\right)\left(x+3\right)}-\frac{3}{2\left(x-3\right)}-1=$
$=\frac{2\left(x^2+x-3\right)-3\left(x+3\right)-2\left(x^2-9\right)}{2\left(x-3\right)\left(x+3\right)}=$
$=\frac{2x^2+2x-6-3x-9-2x^2+18}{2\left(x^2-9\right)}=$
$= \frac{-x+3}{2\left(x-3\right)\left(x+3\right)}=$
$=\frac{-(x-3)}{2(x-3)(x+3)}=-\frac{1}{2(x+3)};$
4) $\frac{ax-2x}{3y+6} ∶\frac{4-a^2}{y^2+4y+4}=$
$= \frac{x\left(a-2\right)}{3\left(y+2\right)}\cdot\frac{(y+2)^2}{\left(2-a\right)\left(2+a\right)}=$
$=\frac{x(a-2)(y+2)^2}{-3(y+2)(a-2)(a+2)}=-\frac{x(y+2)}{3(a+2)}.$