151° Вправа 151Алгебра 8 клас Тарасенкова 2025 · § 4 · стор. 48

151. Знайдіть суму дробів.

1) ${\displaystyle \frac{2^{\mathrm{\setminus }4}}{3ab}+\frac{3}{12ab}=}$${\displaystyle \frac{2\cdot 4}{12ab}+\frac{3}{12ab}=}$${\displaystyle \frac{2\cdot 4+3}{12ab}=}$${\displaystyle \frac{11}{12ab}}$;

2) ${\displaystyle \frac{5y}{36ac}+\frac{{4y}^{\mathrm{\setminus }4}}{9ac}=}$${\displaystyle \frac{5y}{36ac}+\frac{4y\cdot 4}{36ac}=}$${\displaystyle \frac{5y+4y\cdot 4}{36ac}=}$${\displaystyle \frac{21y}{36ac}=}$${\displaystyle \frac{\stackrel{7}{\cancel{21}}y}{\underset{12}{\cancel{36}}ac}=}$${\displaystyle \frac{7y}{12ac}}$;

3) ${\displaystyle \frac{x^{\mathrm{\setminus }7}}{4a^{2}b}+\frac{{3x}^{\mathrm{\setminus }2}}{14a^{2}b}=}$${\displaystyle \frac{x\cdot 7}{28a^{2}b}+\frac{3x\cdot 2}{28a^{2}b}=}$${\displaystyle \frac{x\cdot 7+3x\cdot 2}{28a^{2}b}=}$${\displaystyle \frac{13x}{28a^{2}b}}$;

4) ${\displaystyle \frac{{3a}^{\mathrm{\setminus }5}}{5(a+b)}+\frac{6a}{25(a+b)}=}$${\displaystyle \frac{3a\cdot 5}{25(a+b)}+\frac{6a}{25(a+b)}=}$${\displaystyle \frac{3a\cdot 5+6a}{25(a+b)}=}$${\displaystyle \frac{21a}{25\left(a+b\right)}}$;

5) ${\displaystyle \frac{{4y}^{\mathrm{\setminus }5}}{3(a-c)}+\frac{7y}{15(a-c)}=}$${\displaystyle \frac{4y\cdot 5}{15(a-c)}+\frac{7y}{15(a-c)}=}$${\displaystyle \frac{4y\cdot 5+7y}{15(a-c)}=}$${\displaystyle \frac{27y}{15(a-c)}=}$${\displaystyle \frac{9y}{5\left(a-c\right)}}$;

6) ${\displaystyle \frac{{3x}^{\mathrm{\setminus }3}}{8(a-2b)}+\frac{{5x}^{\mathrm{\setminus }4}}{6(a-2b)}=}$${\displaystyle \frac{3x\cdot 3}{24(a-2b)}+\frac{5x\cdot 4}{24(a-2b)}=}$${\displaystyle \frac{3x\cdot 3+5x\cdot 4}{24(a-2b)}=}$${\displaystyle \frac{29x}{24\left(a-2b\right)}}$.