Question

393. 1) аb^2-a-b^3+b;
2) b x^2+2b^2-b^3-2x^2;

МОЖНО 393 И 394. ПЖЖЖЖ. ДАЮ 47 БАЛЛОВ. ОЧЕНЬ СРОЧНО

Answer 1

$393)\ \ ab^2-a-b^3+b=b^2(a-b)-(a-b)=(a-b)(b-1)(b+1)\\\\bx^2+2b^2-b^3-2x^2=b(x^2-b^2)-2(x^2-b^2)=(x-b)(x+b)(b-2)\\\\a^3+a^2-ab-b^3=(a-b)(a^2+ab+b^2)+a(a-b)=(a-b)(a^2+ab+b^2+a)\\\\x^2+xy-4x-4y=x(x+y)-4(x+y)=(x+y)(x-4)\\\\x^3-3y^2+3x^2-xy^2=x^2(x+3)-y^2(3+x)=(x+3)(x^2-y^2)=(x+3)(x-y)(x+y)\\\\9n+m^3-m^2n-9m=9(n-m)-m^2(n-m)=(n-m)(3-m)(3+m)$

$394)\ \ 1)\ \ x^3+3x^2-4x-12=0\ \ ,\ \ \ x^2(x+3)-4(x+3)=0\ \ ,\\\\(x+3)(x^2-4)=0\ \ ,\ \ \ (x+3)(x-2)(x+2)=0\\\\x_1=-3\ ,\ x_2=2\ ,\ x_3=-2\\\\2)\ \ x^3-2x^2-x+2=0\ \ ,\ \ \ x^2(x-2)-(x-2)=0\ \ ,\\\\(x-2)(x^2-1)=0\ \ ,\ \ \ x^2(x-1)(x+1)=0\\\\x_1=0\ ,\ x_2=1\ ,\ x_3=-1\\\\3)\ \ 2x^3-x^2-18x+9=0\ \ ,\ \ \ x^2(2x-1)-9(2x-1)=0\ \ ,\\\\(2x-1)(x^2-9)=0\ \ ,\ \ \ x^2(x-3)(x+3)=0\ \ ,\\\\x_1=0\ ,\ \ x_2=3\ ,\ x_3=-3\\\\4)\ \ y^3-y^2-16y+16=0\ \ ,\ \ \ y^2(y-1)-16(y-1)=0\ \ ,$

$(y-1)(y^2-16)=0\ \ ,\ \ (y-1)(y-4)(y+4)=0\ \ ,\\\\y_1=1\ ,\ y_2=-4\ ,\ y_3=4$