Страница 177 номер 664, ГДЗ по алгебре за 9 класс к учебнику Мерзляка

\(c\dfrac{(x - a)(x - b)}{(c - a)(c - b)} + b\dfrac{(x - a)(x - c)}{(b - a)(b - c)} + a\dfrac{(x - b)(x - c)}{(a - b)(a - c)} = x\)
\(c\dfrac{(x - a)(x - b)}{(a - c)(b - c)} - b\dfrac{(x - a)(x - c)}{(a - b)(b - c)} + a\dfrac{(x - b)(x - c)}{(a - b)(a - c)} = x\)
\(\dfrac{x - a}{b - c} \cdot \left( \dfrac{c(x - b)}{a - c} - \dfrac{b(x - c)}{a - b} \right) + \dfrac{a(x - b)(x - c)}{(a - b)(a - c)} = x\)

Упростим содержание больших скобок:

\(\dfrac{c(x - b)}{a - c} - \dfrac{b(x - c)}{a - b} = \dfrac{c(x - b)(a - b) - b(x - c)(a - c)}{(a - c)(a - b)} =\)
\(= \dfrac{xca - \textcolor{red}{abc} - \textcolor{blue}{xbc} + cb^{2} - xab + \textcolor{red}{abc} + \textcolor{blue}{xbc} - bc^{2}}{(a - c)(a - b)} =\)
\(= \dfrac{xa(c - b) - bc(c - b)}{(a - c)(a - b)} = \dfrac{(xa - bc)(c - b)}{(a - c)(a - b)}\)

Вернемся к уравнению

\(\dfrac{x - a}{b - c} \cdot \dfrac{(xa - bc)(c - b)}{(a - c)(a - b)} + \dfrac{a(x - b)(x - c)}{(a - b)(a - c)} = x\)
\(\dfrac{x - a}{b - c} \cdot \left( - \dfrac{(xa - bc)(b - c)}{(a - c)(a - b)} \right) + \dfrac{a(x - b)(x - c)}{(a - b)(a - c)} = x\)
\(\dfrac{- (x - a)(xa - bc)}{(a - c)(a - b)} + \dfrac{a(x - b)(x - c)}{(a - b)(a - c)} = x\)
\(\dfrac{\textcolor{red}{- x^{2}a} + xa^{2} + xbc - \textcolor{blue}{abc} + \textcolor{red}{ax^{2}} - abx - axc + \textcolor{blue}{abc}}{(a - b)(a - c)} = x\)
\(\dfrac{ax(a - c) - xb(a - c)}{(a - b)(a - c)} = x\)
\(\dfrac{x(a - b)(a - c)}{(a - b)(a - c)} = x\)
\(x = x\)

Ответ: \(x \in ( - \infty; + \infty)\).