Question

Дам 40 баллов за решение

Answer 1

Ответ:

$1)\ \ 3a^2-3b^2=3(a-b)(a+b)\\\\6x^2-24=6(x-2)(x+2)\\\\x^3-9x=x\, (x-3)(x+3)\\\\b^4-625=(b^2-25)(b^2+25)=(b-5)(b+5)(b^2+25)\\\\-3x^2+30x-75=-3(x^2-10x+25)=-3\, (x-5)^2\\\\63y^3-84y^2z+28yz^2=7y\, (9y^2-12yz+4z^2)=7y\, (3y-2z)^2\\\\a^3+a^2-9a-9=a^2(a+1)-9(a+1)=(a+1)(a-3)(a+3)\\\\a^2+2ab+b^2-c^2=(a+b)^2-c^2=(a+b-c)(a+b+c)$

$2)\ \ x^5-16x^3=0\ \ ,\ \ x^3(x-4)(x+4)=0\ \ ,\ \ x_1=0\ ,\ x_2=4\ ,\ x_3=-4\\\\x^3+6x^2+9x=0\ \ ,\ \ x(x+3)^2=0\ \ ,\ \ x_1=0\ ,\ x_2=-3\\\\x^3-6x^2-x+6=0\ \ ,\ \ x^2(x-6)-(x-6)=0\ \ ,\ \ (x-6)(x-1)(x+1)=0\ ,\\x_1=6\ ,\ x_2=1\ ,\ x_3=-1$

$3)\ \ a^4-4b^4=(a^2-2b^2)(a^2+2b^2)=(a-\sqrt2b)(a+\sqrt2b)(a^2+2b^2)$