421 Вправа 421Алгебра 8 клас Бевз 2025 · § 8 · стор. 82

421. Спрости вираз.

а) ${\displaystyle (\frac{a^2+b^2}{2b}-a):(\frac{1}{b}-\frac{1}{a})+\frac{ba}{2}}$;

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

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

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

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

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

б) ${\displaystyle (a+b):(\frac{1}{a}+\frac{1}{b})-ab\cdot (\frac{a}{b}+\frac{b}{a}-2)}$;

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

2) ${\displaystyle (a+b):\frac{a+b}{ab}=}$${\displaystyle (a+b)\cdot \frac{ab}{a+b}=}$${\displaystyle \frac{ab\cancel{\left(a+b\right)}}{\cancel{\left(a+b\right)}}=}$${\displaystyle ab}$;

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

4) ${\displaystyle ab\cdot \frac{{(a-b)}^2}{ab}=}$${\displaystyle \frac{\cancel{a}\cancel{b}{(a-b)}^2}{\cancel{a}\cancel{b}}=}$${\displaystyle {(a-b)}^2}$;

5) ${\displaystyle ab-{(a-b)}^2}$=

=${\displaystyle ab-a^2+2ab-b^2=}$${\displaystyle 3ab-a^2-b^2}$.