331 Вправа 331Алгебра 8 клас Бевз 2025 · § 7 · стор. 68

331. Спрости вираз.

а) ${\displaystyle \frac{x{y}^2-x^3}{x^2+y^2}\cdot (\frac{x}{{(x-y)}^2}-\frac{y}{x^2-y^2})+\frac{y}{x-y}}$;

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

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

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

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

3) ${\displaystyle \frac{x}{y-x}+\frac{y}{x-y}=}$${\displaystyle -\frac{x}{x-y}+\frac{y}{x-y}=}$${\displaystyle \frac{-x+y}{x-y}}$=

=${\displaystyle \frac{-\cancel{\left(x-y\right)}}{\cancel{\left(x-y\right)}}=}$${\displaystyle -1}$;

б) ${\displaystyle \frac{4a^2+16+16a}{a-2}\cdot (\frac{a}{2a-4}-\frac{a^2+4}{2a^2-8}-\frac{2}{a^2+2a})-\frac{a+8}{a}}$;

1) ${\displaystyle \frac{a}{2a-4}-\frac{a^2+4}{2a^2-8}-\frac{2}{a^2+2a}}$=

=${\displaystyle \frac{a^{\mathrm{\setminus }a\left(a+2\right)}}{2(a-2)}-\frac{{a^2+4}^{\mathrm{\setminus }a}}{2(a-2)(a+2)}-\frac{2^{\mathrm{\setminus }2\left(a-2\right)}}{a(a+2)}}$=

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

=${\displaystyle \frac{a^3+2a^2-a^3-4a-4a+8}{2a(a-2)(a+2)}=}$${\displaystyle \frac{2a^2-8a+8}{2a(a-2)(a+2)}}$=

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

2) ${\displaystyle \frac{4a^2+16+16a}{a-2}\cdot \frac{a-2}{a(a+2)}=}$${\displaystyle \frac{(4a^2+16+16a)\cdot (a-2)}{(a-2)\cdot a(a+2)}}$=

=${\displaystyle \frac{4\cancel{\left(a+2\right)}(a+2)\cancel{\left(a-2\right)}}{a\cancel{\left(a+2\right)}\cancel{\left(a-2\right)}}=}$${\displaystyle \frac{4(a+2)}{a}}$;

3) ${\displaystyle \frac{4(a+2)}{a}-\frac{a+8}{a}=}$${\displaystyle \frac{4(a+2)-(a+8)}{a}=}$${\displaystyle \frac{4a+8-a-8}{a}}$=

=${\displaystyle \frac{3\cancel{a}}{\cancel{a}}=}$${\displaystyle 3}$.