334 Вправа 334Алгебра 8 клас Бевз 2025 · § 7 · стор. 68

334. Спрости вираз.

а) ${\displaystyle \frac{\frac{a}{n}+\frac{c}{n}}{\frac{a}{n}}=}$${\displaystyle (\frac{a}{n}+\frac{c}{n}):\frac{a}{n}}$;

1) ${\displaystyle \frac{a}{n}+\frac{c}{n}=}$${\displaystyle \frac{a+c}{n}}$;

2) ${\displaystyle \frac{a+c}{n}:\frac{a}{n}=}$${\displaystyle \frac{a+c}{n}\cdot \frac{n}{a}=}$${\displaystyle \frac{\cancel{n}(a+c)}{a\cancel{n}}=}$${\displaystyle \frac{a+c}{a}}$;

б) ${\displaystyle \frac{\frac{a}{b}+\frac{b}{a}}{\frac{a}{b}-\frac{b}{a}}=}$${\displaystyle (\frac{a}{b}+\frac{b}{a}):(\frac{a}{b}-\frac{b}{a})}$;

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

2) ${\displaystyle \frac{a^{\mathrm{\setminus }a}}{b}-\frac{b^{\mathrm{\setminus }b}}{a}=}$${\displaystyle \frac{a^2-b^2}{ab}=}$${\displaystyle \frac{(a-b)(a+b)}{ab}}$;

3) ${\displaystyle \frac{a^2+b^2}{ab}:\frac{(a-b)(a+b)}{ab}=}$${\displaystyle \frac{a^2+b^2}{ab}\cdot \frac{ab}{(a-b)(a+b)}}$=

=${\displaystyle \frac{\cancel{a}\cancel{b}(a^2+b^2)}{\cancel{a}\cancel{b}(a+b)(a-b)}=}$${\displaystyle \frac{a^2+b^2}{\left(a-b\right)\left(a+b\right)}}$.