338 Вправа 338Алгебра 8 клас Бевз 2025 · § 7 · стор. 69

338. Спрости вираз.

а) ${\displaystyle \frac{\frac{1}{a+c}-\frac{1}{x}}{a+c-x}\cdot (\frac{a}{c}-\frac{c}{a})}$;

1) ${\displaystyle \frac{1^{\mathrm{\setminus }x}}{a+c}-\frac{1^{\mathrm{\setminus }a+c}}{x}=}$${\displaystyle \frac{x-(a+c)}{x(a+c)}}$=

=${\displaystyle \frac{x-a-c}{x(a+c)}=}$${\displaystyle -\frac{a+c-x}{x(a+c)}}$;

2) ${\displaystyle -\frac{a+c-x}{x(a+c)}:(a+c-x)=}$${\displaystyle \frac{-(a+c-x)}{x(a+c)}\cdot \frac{1}{a+c-x}}$=

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

3) ${\displaystyle \frac{a^{\mathrm{\setminus }a}}{c}-\frac{c^{\mathrm{\setminus }c}}{a}=}$${\displaystyle \frac{a^2-c^2}{ac}=}$${\displaystyle \frac{(a-c)(a+c)}{ac}}$;

4) ${\displaystyle -\frac{1}{x(a+c)}\cdot \frac{(a-c)(a+c)}{ac}=}$${\displaystyle \frac{-1\cdot (a-c)(a+c)}{x(a+c)\cdot ac}}$=

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

б) ${\displaystyle \frac{1-\frac{2x}{a}+\frac{x^2}{a^2}}{1-\frac{x}{a}}=}$${\displaystyle (1-\frac{2x}{a}+\frac{x^2}{a^2}):(1-\frac{x}{a})}$;

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

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

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

=${\displaystyle \frac{\cancel{a}\cancel{\left(a-x\right)}(a-x)}{\cancel{a}a\cancel{\left(a-x\right)}}=}$${\displaystyle \frac{a-x}{a}}$.