356 Вправа 356Алгебра 8 клас Бевз 2025 · § 7 · стор. 71

356. Якщо x = ab/(a + b), то (x² − a²)/(x² − b²) · b³/a³ = (a + 2b)/(b + 2a). Доведи.

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

=${\displaystyle \frac{a^2{b}^2-a^4-2a^{3}b-a^2{b}^2}{{(a+b)}^2}=}$${\displaystyle \frac{-a^4-2a^{3}b}{{(a+b)}^2}=}$${\displaystyle -\frac{a^3(a+2b)}{{(a+b)}^2}}$;

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

=${\displaystyle \frac{a^2{b}^2-a^2{b}^2-2a{b}^3-b^4}{{(a+b)}^2}=}$${\displaystyle \frac{-2a{b}^3-b^4}{{(a+b)}^2}=}$${\displaystyle -\frac{b^3(2a+b)}{{(a+b)}^2}}$;

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

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

=${\displaystyle \frac{a^3(a+2b)\cancel{{\left(a+b\right)}^2}}{b^3(2a+b)\cancel{{\left(a+b\right)}^2}}=}$${\displaystyle \frac{a^3(a+2b)}{b^3(2a+b)}}$;

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

твердження доведено.