329 Вправа 329Алгебра 8 клас Бевз 2025 · § 7 · стор. 68

329. Спрости вираз.

а) ${\displaystyle \frac{4ab}{b^2-a^2}:(\frac{1}{b^2-a^2}+\frac{1}{a^2+2ab+b^2})}$;

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

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

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

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

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

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

б) ${\displaystyle \frac{3c-3}{c^2-4}\cdot \frac{c+2}{c^2-2c+1}-\frac{3(c+2)}{c^2-4}}$;

1) ${\displaystyle \frac{3c-3}{c^2-4}\cdot \frac{c+2}{c^2-2c+1}=}$${\displaystyle \frac{(3c-3)\cdot (c+2)}{(c^2-4)\cdot (c^2-2c+1)}}$=

=${\displaystyle \frac{3\cancel{\left(c+2\right)}\cancel{\left(c-1\right)}}{\cancel{\left(c+2\right)}\cancel{\left(c-1\right)}(c-1)(c-2)}=}$${\displaystyle \frac{3}{(c-2)(c-1)}}$;

2) ${\displaystyle \frac{3}{(c-2)(c-1)}-\frac{3(c+2)}{c^2-4}}$=

=${\displaystyle \frac{3^{\mathrm{\setminus }c+2}}{(c-2)(c-1)}-\frac{{3(c+2)}^{\mathrm{\setminus }c-1}}{(c-2)(c+2)}}$=

=${\displaystyle \frac{3(c+2)-3(c+2)(c-1)}{(c-2)(c-1)(c+2)}=}$${\displaystyle \frac{3c+6-3c^2-3c+6}{(c-2)(c-1)(c+2)}}$=

=${\displaystyle \frac{12-3c^2}{(c-2)(c-1)(c+2)}=}$${\displaystyle \frac{-3\cancel{\left(c-2\right)\left(c+2\right)}}{(c-1)\cancel{\left(c-2\right)\left(c+2\right)}}=}$${\displaystyle \frac{3}{1-c}}$.