ГДЗ по Алгебре 8 Класс Номер 1285 Макарычев Миндюк — Подробные Ответы

Функция \(y\) от \(x\) задана формулой \(y = \frac{ax+b}{cx+d}\),
где \(ad — bc \neq 0\).

Пусть значениям аргумента \(x_1, x_2, x_3\) и \(x_4\) соответствуют значения функции \(y_1, y_2, y_3\) и \(y_4\). Докажите, что

\[
\frac{y_3 — y_1}{y_3 — y_2} \cdot \frac{y_4 — y_1}{y_4 — y_2} = \frac{x_3 — x_1}{x_3 — x_2} \cdot \frac{x_4 — x_1}{x_4 — x_2}.
\]

\( \frac{y_3 — y_1}{y_3 — y_2} \cdot \frac{y_4 — y_1}{y_4 — y_2} = \frac{x_3 — x_1}{x_3 — x_2} \cdot \frac{x_4 — x_1}{x_4 — x_2} \)

\( y = \frac{ax + b}{cx + d} \)

\( y_3 — y_1 = \frac{ax_3 + b}{cx_3 + d} — \frac{ax_1 + b}{cx_1 + d} = \)

\( = \frac{(ax_3 + b)(cx_1 + d) — (ax_1 + b)(cx_3 + d)}{(cx_3 + d)(cx_1 + d)} = \)

\( = \frac{acx_1x_3 + adx_3 + bcx_1 + bd — acx_1x_3 — adx_1 — bcx_3 — bd}{(cx_3 + d)(cx_1 + d)} = \)

\( = \frac{ad(x_3 — x_1) — bc(x_3 — x_1)}{(cx_3 + d)(cx_1 + d)} = \frac{(x_3 — x_1)(ad — bc)}{(cx_3 + d)(cx_1 + d)} \)

\( y_3 — y_2 = \frac{ax_3 + b}{cx_3 + d} — \frac{ax_2 + b}{cx_2 + d} = \)

\( = \frac{(ax_3 + b)(cx_2 + d) — (ax_2 + b)(cx_3 + d)}{(cx_3 + d)(cx_2 + d)} = \)

\( = \frac{acx_2x_3 + adx_3 + bcx_2 + bd — acx_2x_3 — adx_2 — bcx_3 — bd}{(cx_3 + d)(cx_2 + d)} = \)

\( = \frac{ad(x_3 — x_2) — bc(x_3 — x_2)}{(cx_3 + d)(cx_2 + d)} = \frac{(x_3 — x_2)(ad — bc)}{(cx_3 + d)(cx_2 + d)} \)

\( y_4 — y_1 = \frac{ax_4 + b}{cx_4 + d} — \frac{ax_1 + b}{cx_1 + d} = \)

\( = \frac{(ax_4 + b)(cx_1 + d) — (ax_1 + b)(cx_4 + d)}{(cx_4 + d)(cx_1 + d)} = \)

\( = \frac{acx_1x_4 + adx_4 + bcx_1 + bd — acx_1x_4 — adx_1 — bcx_4 — bd}{(cx_4 + d)(cx_1 + d)} = \)

\( = \frac{ad(x_4 — x_1) — bc(x_4 — x_1)}{(cx_4 + d)(cx_1 + d)} = \frac{(x_4 — x_1)(ad — bc)}{(cx_4 + d)(cx_1 + d)} \)

\( y_4 — y_2 = \frac{ax_4 + b}{cx_4 + d} — \frac{ax_2 + b}{cx_2 + d} = \)

\( = \frac{(ax_4 + b)(cx_2 + d) — (ax_2 + b)(cx_4 + d)}{(cx_4 + d)(cx_2 + d)} = \)

\( = \frac{acx_2x_4 + adx_4 + bcx_2 + bd — acx_2x_4 — adx_2 — bcx_4 — bd}{(cx_4 + d)(cx_2 + d)} = \)

\( = \frac{ad(x_4 — x_2) — bc(x_4 — x_2)}{(cx_4 + d)(cx_2 + d)} = \frac{(x_4 — x_2)(ad — bc)}{(cx_4 + d)(cx_2 + d)} \)

\( \frac{y_3 — y_1}{y_3 — y_2} \cdot \frac{y_4 — y_1}{y_4 — y_2} = \frac{(y_3 — y_1)(y_4 — y_2)}{(y_3 — y_2)(y_4 — y_1)} = \frac{(x_3 — x_1)(ad — bc)}{(cx_3 + d)(cx_1 + d)} \cdot \frac{(x_4 — x_2)(ad — bc)}{(cx_4 + d)(cx_2 + d)} = \)

\( = \frac{(x_4 — x_2)(ad — bc)}{(cx_4 + d)(cx_2 + d)} \cdot \frac{(x_3 — x_1)(ad — bc)}{(cx_3 + d)(cx_1 + d)} = \frac{(x_3 — x_1)(x_4 — x_2)}{(x_3 — x_2)(x_4 — x_1)} \cdot \frac{(x_4 — x_2)(ad — bc)}{(cx_4 + d)(cx_2 + d)} \cdot \frac{(x_3 — x_1)(ad — bc)}{(cx_3 + d)(cx_1 + d)} = \)

\( = \frac{(x_3 — x_1)(x_4 — x_2)}{(x_3 — x_2)(x_4 — x_1)} \)