Упростите целое выражение:
а) $(a^2 + 1)(a^2 + 1) + (a - 1)(a^2 + 1) - a^2$;
б) $(x^2 - 1)(x^4 + x^2 + 1) - (x^2 - 1)(x^2 - 1)(x^2 - 1)$;
в) $(m + \frac{1}{2})(m^2 - \frac{1}{2}m + \frac{1}{4}) - (\frac{1}{2}m^3 - \frac{1}{2}m^2)$;
г) $(\frac{1}{2}a - 2b)(\frac{1}{4}a^2 + ab + 4b^2) - (\frac{1}{8}a^3 - 8b^3)$;
д) $15x^3y^2 - (5xy - 2)(3x^2y + x)$;
е) $\frac{1}{2}(a + b + c)(a + b - c) - ab$;
ж) (a + 2b)(a + c) − (a − 2b)(a − c).
$(a^2 + 1)(a^2 + 1) + (a - 1)(a^2 + 1) - a^2 = a^4 + a^2 + a^2 + 1 + a^3 - a^2 + a - 1 - a^2 = a^4 + a^3 + a$
$(x^2 - 1)(x^4 + x^2 + 1) - (x^2 - 1)(x^2 - 1)(x^2 - 1) = x^6 + x^4 + x^2 - x^4 - x^2 - 1 - (x^4 - x^2 - x^2 + 1)(x^2 - 1) = x^6 - 1 - (x^4 - 2x^2 + 1)(x^2 - 1) = x^6 - 1 - (x^6 - 2x^4 + x^2 - x^4 + 2x^2 - 1) = x^6 - 1 - (x^6 - 3x^4 + 3x^2 - 1) = x^6 - 1 - x^6 + 3x^4 - 3x^2 + 1 = 3x^4 - 3x^2$
$(m + \frac{1}{2})(m^2 - \frac{1}{2}m + \frac{1}{4}) - (\frac{1}{2}m^3 - \frac{1}{2}m^2) = m^3 - \frac{1}{2}m^2 + \frac{1}{4}m + \frac{1}{2}m^2 - \frac{1}{4}m + \frac{1}{8} - \frac{1}{2}m^3 + \frac{1}{2}m^2 = \frac{1}{2}m^3 + \frac{1}{2}m^2 + \frac{1}{8}$
$(\frac{1}{2}a - 2b)(\frac{1}{4}a^2 + ab + 4b^2) - (\frac{1}{8}a^3 - 8b^3) = \frac{1}{8}a^3 + \frac{1}{2}a^2b + 2ab^2 - \frac{1}{2}a^2b - 2ab^2 - 8b^3 - \frac{1}{8}a^3 + 8b^3 = 0$
$15x^3y^2 - (5xy - 2)(3x^2y + x) = 15x^3y^2 - (15x^3y^2 - 6x^2y + 5x^2y - 2x) = 15x^3y^2 - 15x^3y^2 + 6x^2y - 5x^2y + 2x = x^2y + 2x$
$\frac{1}{2}(a + b + c)(a + b - c) - ab = \frac{1}{2}(a^2 + ab + ac + ab + b^2 + bc - ac - bc - c^2) - ab = \frac{1}{2}(a^2 + 2ab + b^2 - c^2) - ab = \frac{1}{2}a^2 + ab + \frac{1}{2}b^2 - \frac{1}{2}c^2 - ab = \frac{1}{2}a^2 + \frac{1}{2}b^2 - \frac{1}{2}c^2$
$(a + 2b)(a + c) - (a - 2b)(a - c) = a^2 + 2ab + ac + 2bc - (a^2 - 2ab - ac + 2bc) = a^2 + 2ab + ac + 2bc - a^2 + 2ab + ac - 2bc = 4ab + 2ac$
Пожауйста, оцените решение
