271 Вправа 271Алгебра 8 клас Бевз 2025 · § 6 · стор. 58

271. Знайди частку.

а) ${\displaystyle \frac{4c^2-x^2}{3cx}:\frac{2c+x}{6c{x}^2}=}$${\displaystyle \frac{4c^2-x^2}{3cx}\cdot \frac{6c{x}^2}{2c+x}}$=

=${\displaystyle \frac{(4c^2-x^2)\cdot 6c{x}^2}{3cx\cdot (2c+x)}}$=

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

б) ${\displaystyle 2a{c}^2:\frac{a{c}^2}{3a-c}=}$${\displaystyle 2a{c}^2\cdot \frac{3a-c}{a{c}^2}}$=

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

в) ${\displaystyle \frac{x^3-3x}{x^2-4}:\frac{x^2-3}{x^2+2x}=}$${\displaystyle \frac{x^3-3x}{x^2-4}\cdot \frac{x^2+2x}{x^2-3}}$=

=${\displaystyle \frac{(x^3-3x)\cdot (x^2+2x)}{(x^2-4)\cdot (x^2-3)}=}$${\displaystyle \frac{x^2\cancel{\left(x+2\right)}\cancel{\left(x^2-3\right)}}{\cancel{\left(x+2\right)}\cancel{\left(x^2-3\right)}(x-2)}=}$${\displaystyle \frac{x^2}{x-2}}$.