328 Вправа 328Алгебра 8 клас Бевз 2025 · § 7 · стор. 68

328. Спрости вираз.

а) ${\displaystyle (\frac{3}{a-1}-\frac{3a^2+6a+3}{a^2-1}:\frac{a^4+a}{a^3+1})\cdot \frac{a-a^2}{3}}$;

1) ${\displaystyle \frac{3a^2+6a+3}{a^2-1}:\frac{a^4+a}{a^3+1}=}$${\displaystyle \frac{3a^2+6a+3}{a^2-1}\cdot \frac{a^3+1}{a^4+a}}$=

=${\displaystyle \frac{(3a^2+6a+3)\cdot (a^3+1)}{(a^2-1)\cdot (a^4+a)}}$=

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

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

=${\displaystyle \frac{-3}{a(a-1)}=}$${\displaystyle \frac{3}{a(1-a)}}$;

3) ${\displaystyle \frac{3}{a(1-a)}\cdot \frac{a-a^2}{3}=}$${\displaystyle \frac{3\cdot (a-a^2)}{a(1-a)\cdot 3}=}$${\displaystyle \frac{\cancel{3}\cancel{a}\cancel{\left(1-a\right)}}{\cancel{3}\cancel{a}\cancel{\left(1-a\right)}}=}$${\displaystyle 1}$;

б) ${\displaystyle (\frac{4a}{a+2}-\frac{a^3-8}{a^3+8}:\frac{a^2-4}{4a^2-8a+16})\cdot \frac{a+2}{16}}$;

1) ${\displaystyle \frac{a^3-8}{a^3+8}:\frac{a^2-4}{4a^2-8a+16}=}$${\displaystyle \frac{a^3-8}{a^3+8}\cdot \frac{4a^2-8a+16}{a^2-4}}$=

=${\displaystyle \frac{(a^3-8)\cdot (4a^2-8a+16)}{(a^3+8)\cdot (a^2-4)}}$=

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

2) ${\displaystyle \frac{{4a}^{\mathrm{\setminus }a+2}}{a+2}-\frac{4(a^2+2a+4)}{{(a+2)}^2}=}$${\displaystyle \frac{4a(a+2)-4(a^2+2a+4)}{{(a+2)}^2}}$=

=${\displaystyle \frac{4a^2+8a-4a^2-8a-16}{{(a+2)}^2}=}$${\displaystyle \frac{-16}{{(a+2)}^2}=}$${\displaystyle -\frac{16}{{(a+2)}^2}}$;

3) ${\displaystyle -\frac{16}{{(a+2)}^2}\cdot \frac{a+2}{16}=}$${\displaystyle \frac{-16\cdot (a+2)}{{(a+2)}^2\cdot 16}}$=

=${\displaystyle -\frac{\cancel{16}\cancel{\left(a+2\right)}}{\cancel{16}\cancel{\left(a+2\right)}(a+2)}=}$${\displaystyle -\frac{1}{a+2}}$.