Дана функция y = f(x), где $f(x) = x^2$. Найдите:
а) $f(-12) - 44, f(9) - 1, f(7) - f(3), f(3) + f(4)$;
б) $f(a + b), f(a) + b, f(b) - a, f(a) + f(b)$;
в) $f(ab), af(b), -bf(a), f(\frac{a}{b})$;
г) $f(x - 1) + f(x + 1), f(x + 2) - f(x), \frac{f(x) - 1}{f(x - 1)}, \frac{f(x + 2)}{f(x) - 4}$.
$f(-12) - 44 = (-12)^2 - 44 = 144 - 44 = 100$
$f(9) - 1 = 9^2 - 1 = 81 - 1 = 80$
$f(7) - f(3) = 7^2 - 3^2 = 49 - 9 = 40$
$f(3) + f(4) = 3^2 + 4^2 = 9 + 16 = 25$
$f(a + b) = (a + b)^2 = a^2 + 2ab + b^2$
$f(a) + b = a^2 + b$
$f(b) - a = b^2 - a$
$f(a) + f(b) = a^2 + b^2$
$f(ab) = (ab)^2 = a^2b^2$
$af(b) = ab^2$
$-bf(a) = -ba^2 = -a^2b$
$f(\frac{a}{b}) = (\frac{a}{b})^2 = \frac{a^2}{b^2}$
$f(x - 1) + f(x + 1) = (x - 1)^2 + (x + 1)^2 = x^2 - 2x + 1 + x^2 + 2x + 1 = 2x^2 + 2$
$f(x + 2) - f(x) = (x + 2)^2 - x^2 = x^2 + 4x + 4 - x^2 = 4x + 4$
$\frac{f(x) - 1}{f(x - 1)} = \frac{x^2 - 1}{(x - 1)^2} = \frac{(x - 1)(x + 1)}{(x - 1)^2} = \frac{x + 1}{x - 1}$
$\frac{f(x + 2)}{f(x) - 4} = \frac{(x + 2)^2}{x^2 - 4} = \frac{(x + 2)^2}{(x - 2)(x + 2)} = \frac{x + 2}{x - 2}$
Пожауйста, оцените решение
