150° Вправа 150Алгебра 8 клас Тарасенкова 2025 · § 4 · стор. 47

150. Спростіть вираз.

1) ${\displaystyle \frac{2^{\mathrm{\setminus }2}}{5x}+\frac{3}{10x}=}$${\displaystyle \frac{2\cdot 2}{10x}+\frac{3}{10x}=}$${\displaystyle \frac{2\cdot 2+3}{10x}=}$${\displaystyle \frac{4+3}{10x}=}$${\displaystyle \frac{7}{10x}}$;

2) ${\displaystyle \frac{5^{\mathrm{\setminus }2}}{3a}+\frac{5}{6a}=}$${\displaystyle \frac{5\cdot 2}{6a}+\frac{5}{6a}=}$${\displaystyle \frac{5\cdot 2+5}{6a}=}$${\displaystyle \frac{15}{6a}=}$${\displaystyle \frac{\stackrel{5}{\cancel{15}}}{\underset{2}{\cancel{6}}a}=}$${\displaystyle \frac{5}{2a}}$;

3) ${\displaystyle \frac{5^{\mathrm{\setminus }3}}{12c}+\frac{7^{\mathrm{\setminus }4}}{9c}=}$${\displaystyle \frac{5\cdot 3}{36c}+\frac{7\cdot 4}{36c}=}$${\displaystyle \frac{5\cdot 3+7\cdot 4}{36c}=}$${\displaystyle \frac{15+28}{36c}=}$${\displaystyle \frac{43}{36c}}$;

4) ${\displaystyle \frac{{3x}^{\mathrm{\setminus }3}}{x+2}+\frac{2x}{3(x+2)}=}$${\displaystyle \frac{3x\cdot 3}{3(x+2)}+\frac{2x}{3(x+2)}=}$${\displaystyle \frac{3x\cdot 3+2x}{3(x+2)}=}$${\displaystyle \frac{11x}{3\left(x+2\right)}}$;

5) ${\displaystyle \frac{y}{5(y+1)}+\frac{{2y}^{\mathrm{\setminus }5}}{y+1}=}$${\displaystyle \frac{y}{5(y+1)}+\frac{2y\cdot 5}{5(y+1)}=}$${\displaystyle \frac{y+2y\cdot 5}{5(y+1)}=}$${\displaystyle \frac{11y}{5\left(y+1\right)}}$;

6) ${\displaystyle \frac{{3c}^{\mathrm{\setminus }3}}{2(c-1)}+\frac{{2c}^{\mathrm{\setminus }2}}{3(c-1)}=}$${\displaystyle \frac{3c\cdot 3}{6(c-1)}+\frac{2c\cdot 2}{6(c-1)}=}$${\displaystyle \frac{3c\cdot 3+2c\cdot 2}{6(c-1)}=}$${\displaystyle \frac{9c+4c}{6(c-1)}=}$${\displaystyle \frac{13c}{6\left(c-1\right)}}$;

7) ${\displaystyle \frac{1^{\mathrm{\setminus }2}}{6x}-\frac{5}{12x}=}$${\displaystyle \frac{1\cdot 2}{12x}-\frac{5}{12x}=}$${\displaystyle \frac{1\cdot 2-5}{12x}=}$${\displaystyle \frac{-3}{12x}=}$${\displaystyle -\frac{1}{4x}}$;

8) ${\displaystyle \frac{{13}^{\mathrm{\setminus }3}}{15c}-\frac{5^{\mathrm{\setminus }5}}{9c}=}$${\displaystyle \frac{13\cdot 3}{45c}-\frac{5\cdot 5}{45c}=}$${\displaystyle \frac{13\cdot 3-5\cdot 5}{45c}=}$${\displaystyle \frac{39-25}{45c}=}$${\displaystyle \frac{14}{45c}}$;

9) ${\displaystyle \frac{{2a}^{\mathrm{\setminus }5}}{x+2}-\frac{4a}{5(x+2)}=}$${\displaystyle \frac{2a\cdot 5}{5(x+2)}-\frac{4a}{5(x+2)}=}$${\displaystyle \frac{2a\cdot 5-4a}{5(x+2)}=}$${\displaystyle \frac{6a}{5\left(x+2\right)}}$;

10) ${\displaystyle \frac{c}{3(y-1)}-\frac{{6c}^{\mathrm{\setminus }3}}{y-1}=}$${\displaystyle \frac{c}{3(y-1)}-\frac{6c\cdot 3}{3(y-1)}=}$${\displaystyle \frac{c-6c\cdot 3}{3(y-1)}=}$${\displaystyle -\frac{17c}{3\left(y-1\right)}}$.