203 Вправа 203Алгебра 8 клас Тарасенкова 2025 · § 5 · стор. 59

203. Спростіть вираз.

1) ${\displaystyle \frac{14a^2{b}^2}{5xy}\cdot \frac{10x^2{y}^3}{21a^2{b}^3}\cdot \frac{3b^2}{4y}=}$${\displaystyle \frac{14\cdot 10\cdot 3\cdot a^2{b}^4{x}^2{y}^3}{5\cdot 21\cdot 4\cdot a^2{b}^{3}x{y}^2}=}$${\displaystyle \frac{420\cancel{a^2{b}^{3}x{y}^2}\cdot bxy}{420\cancel{a^2{b}^{3}x{y}^2}}=}$${\displaystyle bxy}$;

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