8 Вправа 8 (повторення)Алгебра 8 клас Тарасенкова 2025 · Повторення · стор. 317

8. Виконайте дії:

1) ${\displaystyle \frac{1}{4x-x^2}+(\frac{1}{x^2-4x}+\frac{2}{16-x^2}+\frac{1}{16+4x})\cdot {\left(\frac{x+4}{x-4}\right)}^2=}$${\displaystyle \frac{1}{x(4-x)}+(\frac{1}{x(x-4)}+\frac{2}{(4-x)(4+x)}+\frac{1}{4(4+x)})\cdot \frac{{(x+4)}^2}{{(x-4)}^2}=}$${\displaystyle \frac{1}{x(4-x)}+(\frac{{-1}^{\mathrm{\setminus }4\left(4+x\right)}}{x(4-x)}+\frac{2^{\mathrm{\setminus }4x}}{(4-x)(4+x)}+\frac{1^{\mathrm{\setminus }x\left(4-x\right)}}{4(4+x)})\cdot \frac{{(x+4)}^2}{{(x-4)}^2}=}$${\displaystyle \frac{1}{x(4-x)}+\frac{-4(4+x)+8x+x(4-x)}{4x(4-x)(4+x)}\cdot \frac{{(x+4)}^2}{{(x-4)}^2}=}$${\displaystyle \frac{1}{x(4-x)}+\frac{-16-4x+8x+4x-x^2}{4x(4-x)(4+x)}\cdot \frac{{(x+4)}^2}{{(x-4)}^2}=}$${\displaystyle \frac{1}{x(4-x)}+\frac{-(x^2-8x+16)}{4x(4-x)(4+x)}\cdot \frac{{(4+x)}^2}{{(x-4)}^2}=}$${\displaystyle \frac{1}{x(4-x)}-\frac{\cancel{{\left(x-4\right)}^2}\cdot (4+x)\cancel{\left(4+x\right)}}{4x(4-x)\cancel{\left(4+x\right)}\cdot \cancel{{\left(x-4\right)}^2}}=}$${\displaystyle \frac{1^{\mathrm{\setminus }4}}{x(4-x)}-\frac{4+x}{4x(4-x)}=}$${\displaystyle \frac{4-(4+x)}{4x(4-x)}=}$${\displaystyle \frac{\cancel{4}\cancel{-4}-x}{4x(4-x)}=}$${\displaystyle \frac{-\cancel{x}}{4\cancel{x}(4-x)}=}$${\displaystyle \frac{-1}{4(4-x)}=}$${\displaystyle \frac{1}{4\left(x-4\right)}}$;

2) ${\displaystyle \frac{3y^2}{x^4-x{y}^3}+\frac{y}{x^3+x^{2}y+x{y}^2}-\frac{1}{x^2-xy}=}$${\displaystyle \frac{3y^2}{x(x^3-y^3)}+\frac{y}{x(x^2+xy+y^2)}-\frac{1}{x(x-y)}=}$${\displaystyle \frac{3y^2}{x(x-y)(x^2+xy+y^2)}+\frac{y^{\mathrm{\setminus }x-y}}{x(x^2+xy+y^2)}-\frac{1^{\mathrm{\setminus }{x}^2+xy+y^2}}{x(x-y)}=}$${\displaystyle \frac{3y^2+y(x-y)-(x^2+xy+y^2)}{x(x-y)(x^2+xy+y^2)}=}$${\displaystyle \frac{3y^2+xy-y^2-x^2-xy-y^2}{x(x-y)(x^2+xy+y^2)}=}$${\displaystyle \frac{y^2-x^2}{x(x-y)(x^2+xy+y^2)}=}$${\displaystyle \frac{-\cancel{\left(x-y\right)}(x+y)}{x\cancel{\left(x-y\right)}(x^2+xy+y^2)}=}$${\displaystyle -\frac{x+y}{x(x^2+xy+y^2)}=}$${\displaystyle -\frac{x+y}{x^3+x^{2}y+x{y}^2}}$.