71 Вправа 71Алгебра 8 клас Бевз 2025 · § 2 · стор. 20

71. Зведи до спільного знаменника дроби.

а) ${\displaystyle \frac{1}{m}}$ і ${\displaystyle \frac{3}{2n}}$; спільний знаменник ${\displaystyle 2mn}$;

${\displaystyle \frac{1}{m}=}$${\displaystyle \frac{1\cdot 2n}{m\cdot 2n}=}$${\displaystyle \frac{2n}{2mn}}$;

${\displaystyle \frac{3}{2n}=}$${\displaystyle \frac{3\cdot m}{2n\cdot m}=}$${\displaystyle \frac{3m}{2mn}}$;

б) ${\displaystyle \frac{3}{2x}}$ і ${\displaystyle \frac{5}{7x^2}}$; спільний знаменник ${\displaystyle 14x^2}$;

${\displaystyle \frac{3}{2x}=}$${\displaystyle \frac{3\cdot 7x}{2x\cdot 7x}=}$${\displaystyle \frac{21x}{14x^2}}$;

${\displaystyle \frac{5}{7x^2}=}$${\displaystyle \frac{5\cdot 2}{7x^2\cdot 2}=}$${\displaystyle \frac{10}{14x^2}}$;

в) ${\displaystyle \frac{3}{x}}$ і ${\displaystyle \frac{2}{x+y}}$; спільний знаменник ${\displaystyle x(x+y)}$;

${\displaystyle \frac{3}{x}=}$${\displaystyle \frac{3\cdot \left(x+y\right)}{x\cdot \left(x+y\right)}=}$${\displaystyle \frac{3\left(x+y\right)}{x\left(x+y\right)}}$;

${\displaystyle \frac{2}{x+y}=}$${\displaystyle \frac{2\cdot x}{(x+y)\cdot x}=}$${\displaystyle \frac{2x}{x\left(x+y\right)}}$;

г) ${\displaystyle \frac{17}{m}}$ і ${\displaystyle \frac{6}{m(m+n)}}$; спільний знаменник ${\displaystyle m(m+n)}$;

${\displaystyle \frac{17}{m}=}$${\displaystyle \frac{17\cdot \left(m+n\right)}{m\cdot \left(m+n\right)}=}$${\displaystyle \frac{17\left(m+n\right)}{m\left(m+n\right)}}$;

${\displaystyle \frac{6}{m(m+n)}=}$${\displaystyle \frac{6}{m\left(m+n\right)}}$;

ґ) ${\displaystyle \frac{2}{x-3}}$ і ${\displaystyle \frac{2}{x+3}}$; спільний знаменник ${\displaystyle (x-3)(x+3)}$;

${\displaystyle \frac{2}{x-3}=}$${\displaystyle \frac{2\cdot \left(x+3\right)}{(x-3)\cdot \left(x+3\right)}=}$${\displaystyle \frac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}}$;

${\displaystyle \frac{2}{x+3}=}$${\displaystyle \frac{2\cdot \left(x-3\right)}{(x+3)\cdot \left(x-3\right)}=}$${\displaystyle \frac{2\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}}$;

д) ${\displaystyle \frac{5}{x+a}}$ і ${\displaystyle \frac{5}{{(x+a)}^2}}$; спільний знаменник ${\displaystyle {(x+a)}^2}$;

${\displaystyle \frac{5}{x+a}=}$${\displaystyle \frac{5\cdot \left(x+a\right)}{(x+a)\cdot \left(x+a\right)}=}$${\displaystyle \frac{5\left(x+a\right)}{{\left(x+a\right)}^2}}$;

${\displaystyle \frac{5}{{(x+a)}^2}=}$${\displaystyle \frac{5}{{\left(x+a\right)}^2}}$.