ВПР 1 №28 Алгебра = ВПТ 1 №28 Математика
Виконайте дію:
1. $\frac{2}{3p}-\frac{4}{9p};$
2. $\frac{7x^2}{12m}+\frac{x^2}{m};$
3. $\frac{3x-2y}{12}+\frac{y+x}{6};$
4. $\frac{3a+b}{6}-\frac{4a-b}{8};$
5. $\frac{1}{p^2}-\frac{p-2}{p^3};$
6. $\frac{4a+b}{2a}-\frac{6b-a}{3b}.$
Розв'язок:
1. $\frac{2}{3p}-\frac{4}{9p}=\frac{6-4}{9p}=\frac{2}{9p};$
2. $\frac{7x^2}{12m}+\frac{x^2}{m}=\frac{7x^2+12x^2}{12m}=\frac{19x^2}{12m};$
3. $\frac{3x-2y}{12}+\frac{y+x}{6}=$
$= \frac{\left(3x-2y\right)+2\left(y+x\right)}{12}=$
$= \frac{3x-2y+2y+2x}{12}= \frac{5x}{12};$
4. $\frac{3a+b}{6}-\frac{4a-b}{8}=$
$= \frac{4\left(3a+b\right)-3\left(4a-b\right)}{24}=$
$= \frac{12a+4b-12a+3b}{24}=\frac{7b}{24};$
5. $\frac{1}{p^2}-\frac{p-2}{p^3}=\frac{p-\left(p-2\right)}{p^3}=$
$= \frac{p-p+2}{p^3}=\frac{2}{p^3};$
6. $\frac{4a+b}{2a}-\frac{6b-a}{3b}=$
$= \frac{3b\left(4a+b\right)-2a\left(6b-a\right)}{6ab}=$
$= \frac{12ab+3b^2-12ab+2a^2}{6ab}=\frac{2a^2+3b^2}{6ab}.$