305 Вправа 305Алгебра 8 клас Бевз 2025 · § 7 · стор. 65

305. Виконай дії.

а) ${\displaystyle \frac{x+2}{x+3}:\frac{5x+10}{9-x^2}-\frac{2x-1}{15}}$;

1) ${\displaystyle \frac{x+2}{x+3}:\frac{5x+10}{9-x^2}=}$${\displaystyle \frac{x+2}{x+3}\cdot \frac{9-x^2}{5x+10}=}$${\displaystyle \frac{(x+2)\cdot (9-x^2)}{(x+3)\cdot (5x+10)}}$=

=${\displaystyle -\frac{\cancel{\left(x+2\right)}\cancel{\left(x+3\right)}(x-3)}{5\cancel{\left(x+2\right)}\cancel{\left(x+3\right)}}=}$${\displaystyle \frac{3-x}{5}}$;

2) ${\displaystyle \frac{{3-x}^{\mathrm{\setminus }3}}{5}-\frac{2x-1}{15}=}$${\displaystyle \frac{3(3-x)-(2x-1)}{15}}$=

=${\displaystyle \frac{9-3x-2x+1}{15}=}$${\displaystyle \frac{10-5x}{15}=}$${\displaystyle \frac{(2-x)\cancel{5}}{3\cancel{5}}=}$${\displaystyle \frac{2-x}{3}}$;

б) ${\displaystyle (\frac{6y}{y-4}-3y):\frac{6-y}{y-4}}$;

1) ${\displaystyle \frac{6y}{y-4}-\frac{{3y}^{\mathrm{\setminus }y-4}}{1}=}$${\displaystyle \frac{6y-3y(y-4)}{y-4}=}$${\displaystyle \frac{6y-3y^2+12y}{y-4}}$=

=${\displaystyle \frac{18y-3y^2}{y-4}=}$${\displaystyle \frac{3y(6-y)}{y-4}}$;

2) ${\displaystyle \frac{3y(6-y)}{y-4}:\frac{6-y}{y-4}=}$${\displaystyle \frac{3y(6-y)}{y-4}\cdot \frac{y-4}{6-y}}$=

=${\displaystyle \frac{3y\cancel{\left(y-4\right)}\cancel{\left(6-y\right)}}{\cancel{\left(y-4\right)}\cancel{\left(6-y\right)}}=}$${\displaystyle 3y}$;

в) ${\displaystyle \frac{x^2-y^2}{{(x-y)}^2}\cdot \frac{x^2}{x+y}+\frac{1-x^2}{x-y}}$;

1) ${\displaystyle \frac{x^2-y^2}{{(x-y)}^2}\cdot \frac{x^2}{x+y}=}$${\displaystyle \frac{(x^2-y^2)\cdot x^2}{{(x-y)}^2\cdot (x+y)}}$=

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

2) ${\displaystyle \frac{x^2}{x-y}+\frac{1-x^2}{x-y}=}$${\displaystyle \frac{x^2+1-x^2}{x-y}=}$${\displaystyle \frac{1}{x-y}}$;

г) ${\displaystyle \frac{5-m}{5+m}\cdot (m+\frac{2m^2}{5-m})}$;

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

=${\displaystyle \frac{m^2+5m}{5-m}=}$${\displaystyle \frac{m(m+5)}{5-m}}$;

2) ${\displaystyle \frac{5-m}{5+m}\cdot \frac{m(m+5)}{5-m}=}$${\displaystyle \frac{m\cancel{\left(m+5\right)}\cancel{\left(5-m\right)}}{\cancel{\left(m+5\right)}\cancel{\left(5-m\right)}}=}$${\displaystyle m}$.