ГДЗ по Алгебре 7 Класс Номер 46.23 Мордкович — Подробные Ответы

Дана функция у = f(x), где f(x) = \(х^2\). Найдите: а) f(-12) — 44, f(9) — 1, f(7) — f(3), f(3) + :(4); б) f(а + b), f(а) + b, f(b) — а, f(а) + f(b); в) f(ab), af(b), -bf(a), f(a:b); г) f(x — 1) + f(x + 1), f(x + 2) — f(x), (f(x)-1):f(x-1), (f(x+2)):(f(x)-4)),

\(f(x) = x^2\)

а) \(f(-12) — 44 = (-12)^2 — 44 = 144 — 44 = 100.\)
\(f(9) — 1 = 9^2 — 1 = 81 — 1 = 80.\)
\(f(7) — f(3) = 7^2 — 3^2 = 49 — 9 = 40.\)
\(f(3) + f(4) = 3^2 + 4^2 = 9 + 16 = 25.\)

б) \(f(a + b) = (a + b)^2 = a^2 + 2ab + b^2.\)
\(f(a) + b = a^2 + b.\)
\(f(b) — a = b^2 — a.\)
\(f(a) + f(b) = a^2 + b^2.\)

в) \(f(ab) = (ab)^2 = a^2b^2.\)
\(af(b) = ab^2.\)
\(-bf(a) = -ba^2.\)
\(f\!\left(\frac{a}{b}\right) = \left(\frac{a}{b}\right)^2 = \frac{a^2}{b^2}.\)

г) \(f(x — 1) + f(x + 1) = (x — 1)^2 + (x + 1)^2 =\)
\(= x^2 — 2x + 1 + x^2 + 2x + 1 = 2x^2 + 2.\)
\(f(x + 2) — f(x) = (x + 2)^2 — x^2 = x^2 + 4x + 4 — x^2 = 4x + 4.\)
\(\frac{f(x) — 1}{f(x — 1)} = \frac{x^2 — 1}{(x — 1)^2} = \frac{(x — 1)(x + 1)}{(x — 1)^2} = \frac{x + 1}{x — 1}.\)
\(\frac{f(x + 2)}{f(x) — 4} = \frac{(x + 2)^2}{x^2 — 4} = \frac{(x + 2)^2}{(x — 2)(x + 2)} = \frac{x + 2}{x — 2}.\)

Решение:
Дана функция \(f(x) = x^2\).

а)
Найдем \(f(-12) — 44\):
\(f(-12) = (-12)^2 = 144\)
\(f(-12) — 44 = 144 — 44 = 100\)

Найдем \(f(9) — 1\):
\(f(9) = (9)^2 = 81\)
\(f(9) — 1 = 81 — 1 = 80\)

Найдем \(f(7) — f(3)\):
\(f(7) = (7)^2 = 49\)
\(f(3) = (3)^2 = 9\)
\(f(7) — f(3) = 49 — 9 = 40\)

Найдем \(f(3) + f(4)\):
\(f(3) = (3)^2 = 9\)
\(f(4) = (4)^2 = 16\)
\(f(3) + f(4) = 9 + 16 = 25\)

б)
Найдем \(f(a + b)\):
\(f(a + b) = (a + b)^2\)
\(f(a + b) = a^2 + 2ab + b^2\)

Найдем \(f(a) + b\):
\(f(a) = a^2\)
\(f(a) + b = a^2 + b\)

Найдем \(f(b) — a\):
\(f(b) = b^2\)
\(f(b) — a = b^2 — a\)

Найдем \(f(a) + f(b)\):
\(f(a) = a^2\)
\(f(b) = b^2\)
\(f(a) + f(b) = a^2 + b^2\)

в)
Найдем \(f(ab)\):
\(f(ab) = (ab)^2\)
\(f(ab) = a^2 b^2\)

Найдем \(af(b)\):
\(f(b) = b^2\)
\(af(b) = a \cdot b^2\)
\(af(b) = ab^2\)

Найдем \(-bf(a)\):
\(f(a) = a^2\)
\(-bf(a) = -b \cdot a^2\)
\(-bf(a) = -a^2 b\)

Найдем \(f(a:b)\):
\(f(a:b) = (a:b)^2\)
\(f(a:b) = \frac{a^2}{b^2}\)

г)
Найдем \(f(x — 1) + f(x + 1)\):
\(f(x — 1) = (x — 1)^2 = x^2 — 2x + 1\)
\(f(x + 1) = (x + 1)^2 = x^2 + 2x + 1\)
\(f(x — 1) + f(x + 1) = (x^2 — 2x + 1) + (x^2 + 2x + 1)\)
\(f(x — 1) + f(x + 1) = 2x^2 + 2\)

Найдем \(f(x + 2) — f(x)\):
\(f(x + 2) = (x + 2)^2 = x^2 + 4x + 4\)
\(f(x) = x^2\)
\(f(x + 2) — f(x) = (x^2 + 4x + 4) — x^2\)
\(f(x + 2) — f(x) = 4x + 4\)

Найдем \((f(x) — 1) : f(x — 1)\):
\(f(x) = x^2\)
\(f(x — 1) = (x — 1)^2 = x^2 — 2x + 1\)
\(\frac{f(x) — 1}{f(x — 1)} = \frac{x^2 — 1}{(x — 1)^2}\)
\(\frac{f(x) — 1}{f(x — 1)} = \frac{(x — 1)(x + 1)}{(x — 1)^2}\)
\(\frac{f(x) — 1}{f(x — 1)} = \frac{x + 1}{x — 1}\), при условии \(x \neq 1\)

Найдем \((f(x + 2)) : (f(x) — 4)\):
\(f(x + 2) = (x + 2)^2 = x^2 + 4x + 4\)
\(f(x) = x^2\)
\(\frac{f(x + 2)}{f(x) — 4} = \frac{(x + 2)^2}{x^2 — 4}\)
\(\frac{f(x + 2)}{f(x) — 4} = \frac{(x + 2)^2}{(x — 2)(x + 2)}\)
\(\frac{f(x + 2)}{f(x) — 4} = \frac{x + 2}{x — 2}\), при условии \(x \neq -2\)

Ответы:
а)
\(100, 80, 40, 25\)
б)
\(a^2 + 2ab + b^2, a^2 + b, b^2 — a, a^2 + b^2\)
в)
\(a^2 b^2, ab^2, -a^2 b, \frac{a^2}{b^2}\)
г)
\(2x^2 + 2, 4x + 4, \frac{x + 1}{x — 1}, \frac{x + 2}{x — 2}\)