ВПР 1 №62 Алгебра = ВПТ 3 №25 Математика
Спростіть вираз:
$\left(\frac{4}{x^2-6x}-\frac{2}{6-x}+1\right)\cdot$
$\cdot\left(\frac{1}{x^2-4}-\frac{2}{x^3-6x^2+12x-8}\right).$
Розв'язок:
$\left(\frac{4}{x^2-6x}-\frac{2}{6-x}+1\right)\cdot$
$\cdot\left(\frac{1}{x^2-4}-\frac{2}{x^3-6x^2+12x-8}\right)=$
$= \frac{1}{x^2-4};$
а) $ \frac{4}{x^2-6x}-\frac{2}{6-x}+1=$
$= \frac{4}{x\left(x-6\right)}+\frac{2}{x-6}+\frac{1}{1}=$
$= \frac{4+2x+x^2-6}{x\left(x-6\right)}=\frac{4-2x+x^2-6x}{x\left(x-6\right)}=$
$\frac{x^2-4x-4}{x\left(x-6\right)}= \frac{\left(x-2\right)^2}{x\left(x-6\right)};$
б) $ \frac{1}{x^2-4}-\frac{2}{x^3-6x^2+12x-8}=$
$=\frac{1}{x^2-4}-\frac{2}{\left(x^3-8\right)-\left(6x^2-12x\right)}=$
$= \frac{1}{\left(x-2\right)\left(x+2\right)}-$
$- \frac{2}{\left(x-2\right)\left(x^2+2x+4\right)-6x\left(x-2\right)}=$
$= \frac{1}{\left(x-2\right)\left(x+2\right)}-$
$- \frac{2}{\left(x-2\right)\left(x^2+2x+4-6x\right)}=$
$= \frac{1}{\left(x-2\right)\left(x+2\right)}-$
$- \frac{2}{\left(x-2\right)\left(x^2-4x+4\right)}=$
$= \frac{1}{\left(x-2\right)\left(x+2\right)}-\frac{2}{\left(x-2\right)\left(x-2\right)}=$
$= \frac{1}{\left(x-2\right)\left(x+2\right)}-\frac{2}{\left(x-2\right)^3}=$
$= \frac{\left(x-2\right)^2-2\left(x+2\right)}{\left(x-2\right)^3\left(x+2\right)}=$
$= \frac{x^2-4x+4-2x-4}{\left(x-2\right)^3\left(x+2\right)}=$
$= \frac{x^2-6x}{\left(x-2\right)^3\left(x+2\right)}=\frac{x\left(x-6\right)}{\left(x-2\right)^3\left(x+2\right)};$
в) $\frac{\left(x-2\right)^2}{x\left(x-6\right)}\cdot\frac{x\left(x-6\right)}{\left(x-2\right)^3\left(x+2\right)}=$
$= \frac{1}{\left(x-2\right)\left(x+2\right)}=\frac{1}{x^2-4}.$