122° Вправа 122Алгебра 8 клас Тарасенкова 2025 · § 3 · стор. 39

122. Зведіть до спільного знаменника дроби.

1) ${\displaystyle \frac{1}{ab}}$ і ${\displaystyle \frac{1}{ac}}$;

${\displaystyle \frac{1^{\mathrm{\setminus }c}}{ab}=}$${\displaystyle \frac{1\cdot c}{ab\cdot c}=}$${\displaystyle \frac{c}{abc}}$;

${\displaystyle \frac{1^{\mathrm{\setminus }b}}{ac}=}$${\displaystyle \frac{1\cdot b}{ac\cdot b}=}$${\displaystyle \frac{b}{abc}}$;

2) ${\displaystyle \frac{1}{a{x}^2}}$ і ${\displaystyle \frac{1}{b{x}^2}}$;

${\displaystyle \frac{1^{\mathrm{\setminus }b}}{a{x}^2}=}$${\displaystyle \frac{1\cdot b}{a{x}^2\cdot b}=}$${\displaystyle \frac{b}{ab{x}^2}}$;

${\displaystyle \frac{1^{\mathrm{\setminus }a}}{b{x}^2}=}$${\displaystyle \frac{1\cdot a}{b{x}^2\cdot a}=}$${\displaystyle \frac{a}{ab{x}^2}}$;

3) ${\displaystyle \frac{1}{x^{2}y}}$ і ${\displaystyle \frac{1}{x^{3}y}}$;

${\displaystyle \frac{1^{\mathrm{\setminus }x}}{x^{2}y}=}$${\displaystyle \frac{1\cdot x}{x^{2}y\cdot x}=}$${\displaystyle \frac{x}{x^{3}y}}$;

${\displaystyle \frac{1}{x^{3}y}}$ — без змін;

4) ${\displaystyle \frac{2}{a{b}^3}}$ і ${\displaystyle \frac{3}{a^{3}b}}$;

${\displaystyle \frac{2^{\mathrm{\setminus }{a}^2}}{a{b}^3}=}$${\displaystyle \frac{2\cdot a^2}{a{b}^3\cdot a^2}=}$${\displaystyle \frac{2a^2}{a^3{b}^3}}$;

${\displaystyle \frac{3^{\mathrm{\setminus }{b}^2}}{a^{3}b}=}$${\displaystyle \frac{3\cdot b^2}{a^{3}b\cdot b^2}=}$${\displaystyle \frac{3b^2}{a^3{b}^3}}$;

5) ${\displaystyle \frac{4}{5b^2}}$ і ${\displaystyle \frac{3}{15b}}$;

${\displaystyle \frac{4^{\mathrm{\setminus }3}}{5b^2}=}$${\displaystyle \frac{4\cdot 3}{5b^2\cdot 3}=}$${\displaystyle \frac{12}{15b^2}}$;

${\displaystyle \frac{3^{\mathrm{\setminus }b}}{15b}=}$${\displaystyle \frac{3\cdot b}{15b\cdot b}=}$${\displaystyle \frac{3b}{15b^2}}$;

6) ${\displaystyle \frac{3}{7y^4}}$ і ${\displaystyle \frac{2}{21y^2}}$;

${\displaystyle \frac{3^{\mathrm{\setminus }3}}{7y^4}=}$${\displaystyle \frac{3\cdot 3}{7y^4\cdot 3}=}$${\displaystyle \frac{9}{21y^4}}$;

${\displaystyle \frac{2^{\mathrm{\setminus }{y}^2}}{21y^2}=}$${\displaystyle \frac{2\cdot y^2}{21y^2\cdot y^2}=}$${\displaystyle \frac{2y^2}{21y^4}}$;

7) ${\displaystyle -\frac{5b}{8a^3}}$ і ${\displaystyle -\frac{6b}{16a}}$;

${\displaystyle -\frac{{5b}^{\mathrm{\setminus }2}}{8a^3}=}$${\displaystyle -\frac{5b\cdot 2}{8a^3\cdot 2}=}$${\displaystyle -\frac{10b}{16a^3}}$;

${\displaystyle -\frac{{6b}^{\mathrm{\setminus }{a}^2}}{16a}=}$${\displaystyle -\frac{6b\cdot a^2}{16a\cdot a^2}=}$${\displaystyle -\frac{6a^{2}b}{16a^3}}$;

8) ${\displaystyle \frac{4c}{15a{b}^2}}$ і ${\displaystyle -\frac{1}{20ac}}$;

${\displaystyle \frac{{4c}^{\mathrm{\setminus }4c}}{15a{b}^2}=}$${\displaystyle \frac{4c\cdot 4c}{15a{b}^2\cdot 4c}=}$${\displaystyle \frac{16c^2}{60a{b}^{2}c}}$;

${\displaystyle -\frac{1^{\mathrm{\setminus }3b^2}}{20ac}=}$${\displaystyle -\frac{1\cdot 3b^2}{20ac\cdot 3b^2}=}$${\displaystyle -\frac{3b^2}{60a{b}^{2}c}}$;

9) ${\displaystyle -\frac{x}{2y^3}}$ і ${\displaystyle \frac{3}{14y}}$;

${\displaystyle -\frac{x^{\mathrm{\setminus }7}}{2y^3}=}$${\displaystyle -\frac{x\cdot 7}{2y^3\cdot 7}=}$${\displaystyle -\frac{7x}{14y^3}}$;

${\displaystyle \frac{3^{\mathrm{\setminus }{y}^2}}{14y}=}$${\displaystyle \frac{3\cdot y^2}{14y\cdot y^2}=}$${\displaystyle \frac{3y^2}{14y^3}}$.