307 Вправа 307Алгебра 8 клас Бевз 2025 · § 7 · стор. 65

307. Спрости вираз.

а) ${\displaystyle (\frac{x}{y}-\frac{y}{x})\cdot \frac{xy}{x+y}}$;

1) ${\displaystyle \frac{x^{\mathrm{\setminus }x}}{y}-\frac{y^{\mathrm{\setminus }y}}{x}=}$${\displaystyle \frac{x^2-y^2}{xy}=}$${\displaystyle \frac{(x-y)(x+y)}{xy}}$;

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

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

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

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

=${\displaystyle \frac{\cancel{a}\cancel{b}\cancel{\left(a-b\right)}}{\cancel{a}\cancel{b}(a+b)\cancel{\left(a-b\right)}}=}$${\displaystyle \frac{1}{a+b}}$;

в) ${\displaystyle (1+\frac{a}{b}):(1-\frac{a}{b})}$;

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

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

3) ${\displaystyle \frac{a+b}{b}:\frac{b-a}{b}=}$${\displaystyle \frac{a+b}{b}\cdot \frac{b}{b-a}=}$${\displaystyle \frac{\cancel{b}(a+b)}{\cancel{b}(b-a)}=}$${\displaystyle \frac{a+b}{b-a}}$;

г) ${\displaystyle (\frac{1}{xy}+\frac{1}{y}):(\frac{1}{xy}-\frac{1}{y})}$;

1) ${\displaystyle \frac{1}{xy}+\frac{1^{\mathrm{\setminus }x}}{y}=}$${\displaystyle \frac{1+x}{xy}}$;

2) ${\displaystyle \frac{1}{xy}-\frac{1^{\mathrm{\setminus }x}}{y}=}$${\displaystyle \frac{1-x}{xy}}$;

3) ${\displaystyle \frac{x+1}{xy}:\frac{1-x}{xy}=}$${\displaystyle \frac{x+1}{xy}\cdot \frac{xy}{1-x}=}$${\displaystyle \frac{\cancel{x}\cancel{y}(x+1)}{\cancel{x}\cancel{y}(1-x)}=}$${\displaystyle \frac{x+1}{1-x}}$.