355 Вправа 355Алгебра 8 клас Бевз 2025 · § 7 · стор. 71

355. Доведи твердження.

а) Якщо ${\displaystyle x=}$${\displaystyle \frac{a-b}{a+b}}$, то ${\displaystyle (\frac{a}{b}-x)\cdot \frac{b}{ax+b}=}$${\displaystyle 1}$;

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

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

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

=${\displaystyle \frac{a^2-ab+ab+b^2}{a+b}=}$${\displaystyle \frac{a^2+b^2}{a+b}}$;

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

4) ${\displaystyle \frac{a^2+b^2}{b(a+b)}\cdot \frac{b(a+b)}{a^2+b^2}=}$${\displaystyle \frac{\cancel{b}\cancel{\left(a+b\right)}\cancel{\left(a^2+b^2\right)}}{\cancel{b}\cancel{\left(a+b\right)}\cancel{\left(a^2+b^2\right)}}=}$${\displaystyle 1}$;

твердження доведено;

б) якщо ${\displaystyle x=}$${\displaystyle \frac{ab}{a^2+b^2}}$, то ${\displaystyle \frac{0,5-x}{0,5+x}=}$${\displaystyle \frac{{(a-b)}^2}{{(a+b)}^2}}$;

1) ${\displaystyle 0,5-x=}$${\displaystyle \frac{1^{\mathrm{\setminus }{a}^2+b^2}}{2}-\frac{{ab}^{\mathrm{\setminus }2}}{a^2+b^2}=}$${\displaystyle \frac{a^2+b^2-2ab}{2(a^2+b^2)}}$=

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

2) ${\displaystyle 0,5+x=}$${\displaystyle \frac{1^{\mathrm{\setminus }{a}^2+b^2}}{2}+\frac{{ab}^{\mathrm{\setminus }2}}{a^2+b^2}=}$${\displaystyle \frac{a^2+b^2+2ab}{2(a^2+b^2)}}$=

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

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

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

твердження доведено.