Обчисліть значення виразу (405-406).

Решение

405.

а) $$c^3 - 2c^2 + 3c - 4 - (c^3 - 3c^2 - 5) = c^3 - 2c^2 + 3c - 4 - c^3 + 3c^2 + 5 = (c^3 - c^3) + (-2c^2 + 3c^2) + 3c + (-4 + 5) = c^2 + 3c + 1$$.

При $$c = 2$$: $$2^2 + 3 \times 2 + 1 = 4 + 6 + 1 = 11$$.

б) $$4x^2 - (-2x^3 + 4x^2 - 5) = 4x^2 + 2x^3 - 4x^2 + 5 = 2x^3 + (4x^2 - 4x^2) + 5 = 2x^3 + 5$$.

При $$x = -3$$: $$2 \times (-3)^3 + 5 = 2 \times (-27) + 5 = -54 + 5 = -49$$.

в) $$2p - (1 - p^2 - p^3) - (2p + p^2 - p^3) = 2p - 1 + p^2 + p^3 - 2p - p^2 + p^3 = (2p - 2p) + (p^2 - p^2) + (p^3 + p^3) - 1 = 2p^3 - 1$$.

При $$p = \frac{2}{3}$$: $$2 \times (\frac{2}{3})^3 - 1 = 2 \times \frac{8}{27} - 1 = \frac{16}{27} - 1 = \frac{16 - 27}{27} = -\frac{11}{27}$$.

406.

а) $$x^3 - 3x^2 + 3x - 1 - (3x - 3x^2) = x^3 - 3x^2 + 3x - 1 - 3x + 3x^2 = x^3 + (-3x^2 + 3x^2) + (3x - 3x) - 1 = x^3 - 1$$.

При $$x = 3$$: $$3^3 - 1 = 27 - 1 = 26$$.

б) $$5a^4 - 2a^3 - (4a^4 - 2a^3 + 1) = 5a^4 - 2a^3 - 4a^4 + 2a^3 - 1 = (5a^4 - 4a^4) + (-2a^3 + 2a^3) - 1 = a^4 - 1$$.

При $$a = -2$$: $$(-2)^4 - 1 = 16 - 1 = 15$$.

в) $$a^2 - 2ab + b^2 - (a - b - 3) = a^2 - 2ab + b^2 - a + b + 3$$.

При $$a = 5, b = 4$$: $$5^2 - 2 \times 5 \times 4 + 4^2 - 5 + 4 + 3 = 25 - 40 + 16 - 5 + 4 + 3 = 41 - 45 + 7 = -4 + 7 = 3$$.

г) $$2 + xy - x^2 - (y^2 - 2xy + 4) = 2 + xy - x^2 - y^2 + 2xy - 4 = -x^2 - y^2 + (xy + 2xy) + (2 - 4) = -x^2 - y^2 + 3xy - 2$$.

При $$x = 0.2, y = -0.5$$: $$-(0.2)^2 - (-0.5)^2 + 3 \times 0.2 \times (-0.5) - 2 = -0.04 - 0.25 + 3 \times (-0.1) - 2 = -0.04 - 0.25 - 0.3 - 2 = -0.29 - 2.3 = -2.59$$.

Ответ: 405. а) 11; б) -49; в) $$-11/27$$. 406. а) 26; б) 15; в) 3; г) -2.59.