Question

Срочно Алгебра, упростить выражения Номера 35.16,35.17​

Answer 1

$35.16\\ \\ 1) \quad (4x^3-1)(9x^3+5)-(6x^3-1)^2 = 36x^6+20x^3-9x^3-5-(36x^6-12x^3+1) = \\ \\ = 36x^6+11x^3-5-36x^6+12x^3-1=23x^3-6 \\ \\ 2) \quad (x^4-1)^2-(x^4+4)-(x^4-6) = x^8-2x^4+1-x^4-4-x^4+6=x^8-4x^4+3 \\ \\ 3) \quad (x^7-3)(x^7+7)-(x^7+2)^2 = x^{14}+7x^7-3x^7-21-(x^{14}+4x^7+4)=\\ \\ = x^{14}+4x^7-21-x^{14}-4x^7-4=-25 \\ \\ 4) \quad (x^8+9)(11-x^8)-(x^8+1)^2 = 11x^8-x^{16}+99-9x^8-(x^{16}+2x^8+1)= \\ \\ = -x^{16}+2x^8+99-x^{16}-2x^8-1=-2x^{16}+98$

$35.17 \\ \\ 1) \quad (a^3+b^3)^3 -(a^3-b^3)^3-2b^9 = (a^9+3a^6b^3+3a^3b^6+b^9)-\\ \\ -(a^9-3a^6b^3+3a^3b^6-b^9)-2b^9 = a^9 - a^9+3a^6b^3+3a^6b^3+3a^3b^6-3a^3b^6+\\ \\ +b^9+b^9-2b^9 = 6a^6b^3 \\ \\ 2) \quad (1-a^3b^3)^3-(a^3b^3-1)^3-2 = (1-a^3b^3)^3+(1-a^3b^3)^3-2 = \\ \\ = 2(1-a^3b^3)^3-2 = 2(1-3a^3b^3+3a^6b^6-a^9b^9)-2 =-6a^3b^3+6a^6b^6-2a^9b^9 \\ \\ 3) \quad 3a^4b^4(a^4-b^4)-(a^4-b^4)^2 = 3a^8b^4-3a^4b^8-(a^8-2a^4b^4+b^8) =\\ \\ = 3a^8b^4-3a^4b^8-a^8+2a^4b^4-b^8$

$4) \quad (c^2+d^2)^3-3c^2d^2(c^2+d^2) = c^6+3c^4d^2+3c^2d^4+d^6-3c^4d^2-3c^2d^4 = c^6+d^6$