Упростите выражение:
\({\largeа)}\ \frac{1}{a(a-b)(a-c)}+\frac{1}{b(b-c)(b-a)}+\frac{1}{c(c-a)(c-b)};\)
\({\largeб)}\ \frac{x^2}{(x-y)(x-z)}+\frac{y^2}{(y-x)(y-z)}+\frac{z^2}{(z-x)(z-y)}.\)
Упражнение 237
Источник заимствования: Математика. Алгебра. 8 класс. Базовый уровень. / Ю.Н. Макарычев, Н.Г. Миндюк, К.И. Нешков, С.Б. Суворова – Просвещение, 2023. – 59 c. ISBN 978-5-09-102536-1
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Решение:
\({\largeа)}\ \frac{1}{ a(a-b)(a-c)}+\frac{1}{ b(b-c)(b-a)}+\frac{1}{ c(c-a)(c-b)}=\frac{1}{ a(a-b)(a-c)}-\frac{1}{ b(b-c)(a-b)}+\frac{1}{ c(a-c)(b-c)}=\frac{ bc(b-c)-ac(a-c)+ab(a-b)}{ abc(a-b)(a-c)(b-c)}=\frac{b^2c-bc^2-a^2c+ac^2+a^2b-ab^2}{ abc(a-b)(a-c)(b-c)}=\frac{{-}ab^2+b^2c+a^2b-bc^2-a^2c+ac^2}{ abc(a-b)(a-c)(b-c)}=\frac{{-}b^2(a-c)+b(a^2-c^2)-ac(a-c)}{ abc(a-b)(a-c)(b-c)}=\frac{{-}b^2(a-c)+b(a-c)(a+c)-ac(a-c)}{ abc(a-b)(a-c)(b-c)}=\frac{ (a-c)({-}b^2+b(a+c)-ac)}{ abc(a-b)(a-c)(b-c)}=\frac{{-}b^2+ab+bc-ac}{ abc(a-b)(b-c)}=\frac{ab-b^2-ac+bc}{ abc(a-b)(b-c)}=\frac{ b(a-b)-c(a-b)}{ abc(a-b)(b-c)}=\frac{ (a-b)(b-c)}{ abc(a-b)(b-c)}=\frac{1}{abc};\)
\({\largeб)}\ \frac{x^2}{ (x-y)(x-z)}+\frac{y^2}{ (y-x)(y-z)}+\frac{z^2}{ (z-x)(z-y)}=\frac{x^2}{ (x-y)(x-z)}-\frac{y^2}{ (x-y)(y-z)}+\frac{z^2}{ (x-z)(y-z)}=\frac{ x^2(y-z)-y^2(x-z)+z^2(x-y)}{ (x-y)(x-z)(y-z)}=\frac{x^2y-x^2z-xy^2+y^2z+xz^2-yz^2}{ (x-y)(x-z)(y-z)}=\frac{x^2y-xy^2-x^2z+y^2z+xz^2-yz^2}{ (x-y)(x-z)(y-z)}=\frac{ xy(x-y)-z(x^2-y^2)+z^2(x-y)}{ (x-y)(x-z)(y-z)}=\frac{ xy(x-y)-z(x-y)(x+y)+z^2(x-y)}{ (x-y)(x-z)(y-z)}=\frac{ (x-y)(xy-z(x+y)+z^2)}{ (x-y)(x-z)(y-z)}=\frac{xy-xz-yz+z^2}{ (x-z)(y-z)}=\frac{ x(y-z)-z(y-z)}{ (x-z)(y-z)}=\frac{ (x-z)(y-z)}{ (x-z)(y-z)}=1.\)