359 Вправа 359Алгебра 8 клас Бевз 2025 · § 7 · стор. 71

359. Спрости вираз.

а) ${\displaystyle \frac{1}{(a-b)(b-c)}+\frac{1}{(c-a)(a-b)}+\frac{1}{(b-c)(c-a)}}$;

${\displaystyle \frac{1^{\mathrm{\setminus }c-a}}{(a-b)(b-c)}+\frac{1^{\mathrm{\setminus }b-c}}{(c-a)(a-b)}+\frac{1^{\mathrm{\setminus }a-b}}{(b-c)(c-a)}}$=

=${\displaystyle \frac{c-a+b-c+a-b}{(a-b)(b-c)(c-a)}=}$${\displaystyle 0}$;

б) ${\displaystyle \frac{x^2}{(x-y)(x-z)}+\frac{y^2}{(y-x)(y-z)}+\frac{z^2}{(z-x)(z-y)}}$;

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

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

чисельник:

${\displaystyle x^{2}y-x^{2}z-x{y}^2+y^{2}z+z^2(x-y)}$=

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

=${\displaystyle (x-y)(xy-zx-zy+z^2)}$=

=${\displaystyle (x-y)(x(y-z)-z(y-z))}$=

=${\displaystyle (x-y)(x-z)(y-z)}$;

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