214 Вправа 214Алгебра 8 клас Бевз 2025 · § 4 · стор. 42

214. Доведи тотожність.

а) Кожен дріб — різниця квадратів:

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

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

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

сума: ${\displaystyle \frac{x+y-z}{x+y+z}+\frac{y-x+z}{x+y+z}+\frac{z+x-y}{x+y+z}}$=

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

тотожність доведено;

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

=${\displaystyle \frac{{x^2-yz}^{\mathrm{\setminus }y+z}}{(x-y)(x-z)}-\frac{{y^2+xz}^{\mathrm{\setminus }x-z}}{(x-y)(y+z)}-\frac{{z^2+xy}^{\mathrm{\setminus }x-y}}{(x-z)(y+z)}}$=

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

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

=${\displaystyle \frac{0}{(x-y)(x-z)(y+z)}=}$${\displaystyle 0}$;

${\displaystyle 0=}$${\displaystyle 0}$ — тотожність доведено;

в) ${\displaystyle \frac{(y-b)(z-b)}{b(b-c)}+\frac{(y-c)(z-c)}{c(c-b)}+\frac{yz}{bc}}$=

=${\displaystyle \frac{{(y-b)(z-b)}^{\mathrm{\setminus }c}}{b(b-c)}-\frac{{(y-c)(z-c)}^{\mathrm{\setminus }b}}{c(b-c)}+\frac{{yz}^{\mathrm{\setminus }b-c}}{bc}}$=

=${\displaystyle \frac{-c(y-b)(b-z)-b(y-c)(z-c)+yz(b-c)}{bc(b-c)}}$=

=${\displaystyle \frac{b^{2}c-bcy-bcz+cyz-b{c}^2+bcy+bcz-byz+byz-cyz}{bc(b-c)}}$=

=${\displaystyle \frac{b^{2}c-b{c}^2}{bc(b-c)}=}$${\displaystyle \frac{\cancel{bc\left(b-c\right)}}{\cancel{bc\left(b-c\right)}}=}$${\displaystyle 1}$;

${\displaystyle 1=}$${\displaystyle 1}$ — тотожність доведено.