306 Вправа 306Алгебра 8 клас Бевз 2025 · § 7 · стор. 65

306. Виконай дії.

а) ${\displaystyle \frac{a^2-4}{9-a^2}:\frac{a-2}{3+a}-\frac{2}{3-a}}$;

1) ${\displaystyle \frac{a^2-4}{9-a^2}:\frac{a-2}{3+a}=}$${\displaystyle \frac{a^2-4}{9-a^2}\cdot \frac{3+a}{a-2}=}$${\displaystyle \frac{(a^2-4)\cdot (3+a)}{(9-a^2)\cdot (a-2)}}$=

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

2) ${\displaystyle \frac{a+2}{3-a}-\frac{2}{3-a}=}$${\displaystyle \frac{a+2-2}{3-a}=}$${\displaystyle \frac{a}{3-a}}$;

б) ${\displaystyle (x-\frac{5x}{x+3}):\frac{x-2}{x+3}}$;

1) ${\displaystyle \frac{x^{\mathrm{\setminus }x+3}}{1}-\frac{5x}{x+3}=}$${\displaystyle \frac{x(x+3)-5x}{x+3}=}$${\displaystyle \frac{x^2+3x-5x}{x+3}}$=

=${\displaystyle \frac{x^2-2x}{x+3}=}$${\displaystyle \frac{x(x-2)}{x+3}}$;

2) ${\displaystyle \frac{x(x-2)}{x+3}:\frac{x-2}{x+3}=}$${\displaystyle \frac{x(x-2)}{x+3}\cdot \frac{x+3}{x-2}}$=

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

в) ${\displaystyle \frac{a^2-b^2}{4a+4}\cdot \frac{a+1}{a-b}+\frac{a+b}{4}}$;

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

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

2) ${\displaystyle \frac{a+b}{4}+\frac{a+b}{4}=}$${\displaystyle \frac{a+b+a+b}{4}=}$${\displaystyle \frac{2a+2b}{4}}$=

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

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

1) ${\displaystyle \frac{2a}{2a-1}+\frac{1^{\mathrm{\setminus }2a-1}}{1}=}$${\displaystyle \frac{2a+2a-1}{2a-1}=}$${\displaystyle \frac{4a-1}{2a-1}}$;

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

=${\displaystyle \frac{3\cancel{\left(2a-1\right)}\cancel{\left(4a-1\right)}}{a\cancel{\left(2a-1\right)}\cancel{\left(4a-1\right)}}=}$${\displaystyle \frac{3}{a}}$.