№ 7 ЗПС Алгебра = № 7 ЗПС Математика
Доведіть, що коли x + y = 1, то
$\frac{x}{y^3-1}-\frac{y}{x^3-1}=\frac{2(y-x)}{x^2y^2+3}.$
Розв'язок:
$\frac{x}{y^3-1}-\frac{y}{x^3-1}= \frac{x}{(y-1)\left(y^2+y+1\right)}-$
$- \frac{y}{(x-1)\left(x^2+x+1\right)}.$
Оскільки $x+y=1,$
то $y=1-x,x=1-y.$
Отже, $\frac{x}{y^3-1}-\frac{y}{x^3-1}=$
$= \frac{x}{-x\left(1-2x+x^2+1-x+1\right)}-$
$-\frac{1-x}{(x-1)\left(x^2+x+1\right)}=$
$= -\frac{1}{x^2-3x+2}+\frac{1}{x^2+x+1}=$
$= \frac{x^2-3x+3-\left(x^2+x+1\right)}{\left(x^2+x+1\right)\left(x^2-3x+3\right)}=$
$= \frac{x^2-3x+3-x^2-x-1}{\left(x^2+x+1\right)\left(x^2-3x+3\right)}=$
$= \frac{2-4x}{\left(x^2+x+1\right)\left(x^2-3x+3\right)}=$
$= \frac{2(1-2x)}{\left((1-y)^2+1-y+1\right)\left(x^2-3(x-1)\right)}=$
$= \frac{2(1-x-x)}{\left(1+y^2-2y-y+2\right)\left(x^2+3y\right)}=$
$= \frac{2(y-x)}{\left(y^2-3y+3\right)\left(x^2+3y\right)}=$
$= \frac{2(y-x)}{\left(y^2-3(y-1)\right)\left(x^2+3y\right)}=$
$= \frac{2(y-x)}{\left(y^2+3x\right)\left(x^2+3y\right)}=$
$= \frac{2(y-x)}{x^2y^2+3y^3+3x^3+9xy}=$
$= \frac{2(y-x)}{x^2y^2+3(x+y)\left(x^2-xy+y^2\right)+9xy}=$
$= \frac{2(y-x)}{x^2y^2+3\left(x^2-xy+y^2+3xy\right)}=$
$= \frac{2(y-x)}{x^2y^2+3\left(x^2+2xy+y^2\right)}=$
$= \frac{2(y-x)}{x^2y^2+3(x+y)^2}=\frac{2(y-x)}{x^2y^2+3},$
оскільки $x+y=1.$