№ 4.28 Алгебра = № 4.28 Математика
Спростіть вираз:
1) $\frac{a-2}{ab-a^2}-\frac{2-b}{ab-b^2};$
2) $\frac{t^2}{ta+a^2}-\frac{a}{t+a};$
3)$\ \frac{4}{a^2-9}-\frac{2}{a^2+3a}; $
4) $\frac{3n^2-8m^2}{n^2-2mn}-\frac{3mn-n^2}{mn-2m^2}.$
Розв'язок:
1) $\frac{a-2}{ab-a^2}-\frac{2-b}{ab-b^2}=$
$= \frac{a-2}{a\left(b-a\right)}+\frac{2-b}{b\left(b-a\right)}=$
$\frac{b(a-2)+a(2-b)}{ab\left(b-a\right)}=$
$= \frac{ab-2b+2a-ab}{ab\left(b-a\right)}=\frac{2a-2b}{ab\left(b-a\right)}=$
$= \frac{2\left(a-b\right)}{-ab\left(a-b\right)}=-\frac{2}{ab};$
2) $\frac{t^2}{ta+a^2}-\frac{a}{t+a}=\frac{t^2}{a\left(t+a\right)}-\frac{a}{t+a}=$
$= \frac{t^2-a^2}{a\left(t+a\right)}=\frac{\left(t-a\right)\left(t+a\right)}{a\left(t+a\right)}=\frac{t-a}{a};$
3)$\frac{4}{a^2-9}-\frac{2}{a^2+3a}=$
$= \frac{4}{\left(a-3\right)\left(a+3\right)}-\frac{2}{a\left(a+3\right)}=$
$= \frac{4a-2(a-3)}{a\left(a-3\right)\left(a+3\right)}=$
$= \frac{4a-2a+6}{a\left(a-3\right)\left(a+3\right)}=$
$= \frac{2a+6}{a\left(a-3\right)\left(a+3\right)}=$
$= \frac{2(a+3)}{a\left(a-3\right)\left(a+3\right)}=\frac{2}{a^2-3a}; $
4) $\frac{3n^2-8m^2}{n^2-2mn}-\frac{3mn-n^2}{mn-2m^2}=$
$= \frac{3n^2-8m^2}{n(n-2m)}-\frac{3mn-n^2}{m(n-2m)}=$
$= \frac{m(3n^2-8m^2)-n(3mn-n^2)}{mn\left(n-2m\right)}=$
$= \frac{3n^2m-8m^3-3mn^2+n^3}{mn\left(n-2m\right)}=$
$= \frac{n^3-8m^3}{mn\left(n-2m\right)}=$
$= \frac{\left(n-2m\right)\left(n^2+2mn+4m^2\right)}{mn\left(n-2m\right)}=$
$= \frac{n^2+2mn+4m^2}{mn}.$