153° Вправа 153Алгебра 8 клас Тарасенкова 2025 · § 4 · стор. 48

153. Спростіть вираз.

1) ${\displaystyle \frac{1^{\mathrm{\setminus }2}}{8y}+\frac{3}{16y}=}$${\displaystyle \frac{1\cdot 2}{16y}+\frac{3}{16y}=}$${\displaystyle \frac{1\cdot 2+3}{16y}=}$${\displaystyle \frac{5}{16y}}$;

2) ${\displaystyle \frac{{2a}^{\mathrm{\setminus }4}}{a-1}+\frac{5a}{4(a-1)}=}$${\displaystyle \frac{2a\cdot 4}{4(a-1)}+\frac{5a}{4(a-1)}=}$${\displaystyle \frac{2a\cdot 4+5a}{4(a-1)}=}$${\displaystyle \frac{13a}{4\left(a-1\right)}}$;

3) ${\displaystyle \frac{2b}{3(a^2+3)}+\frac{{4b}^{\mathrm{\setminus }3}}{a^2+3}=}$${\displaystyle \frac{2b}{3(a^2+3)}+\frac{4b\cdot 3}{3(a^2+3)}=}$${\displaystyle \frac{2b+4b\cdot 3}{3(a^2+3)}=}$${\displaystyle \frac{14b}{3\left(a^2+3\right)}}$;

4) ${\displaystyle \frac{{ac}^{\mathrm{\setminus }3}}{7x^2}-\frac{2ac}{21x^2}=}$${\displaystyle \frac{ac\cdot 3}{21x^2}-\frac{2ac}{21x^2}=}$${\displaystyle \frac{ac\cdot 3-2ac}{21x^2}=}$${\displaystyle \frac{ac}{21x^2}}$;

5) ${\displaystyle \frac{{2c}^{\mathrm{\setminus }4}}{3(2-a)}-\frac{{5c}^{\mathrm{\setminus }3}}{4(2-a)}=}$${\displaystyle \frac{2c\cdot 4}{12(2-a)}-\frac{5c\cdot 3}{12(2-a)}=}$${\displaystyle \frac{2c\cdot 4-5c\cdot 3}{12(2-a)}=}$${\displaystyle \frac{8c-15c}{12(2-a)}=}$${\displaystyle -\frac{7c}{12(2-a)}=}$${\displaystyle \frac{7c}{12\left(a-2\right)}}$;

6) ${\displaystyle \frac{{3c}^{\mathrm{\setminus }3}}{10(1+x)}-\frac{{2c}^{\mathrm{\setminus }2}}{15(1+x)}=}$${\displaystyle \frac{3c\cdot 3}{30(1+x)}-\frac{2c\cdot 2}{30(1+x)}=}$${\displaystyle \frac{3c\cdot 3-2c\cdot 2}{30(1+x)}=}$${\displaystyle \frac{5c}{30(1+x)}=}$${\displaystyle \frac{c}{6\left(1+x\right)}}$.