Упр.477 ГДЗ Никольский Потапов 7 класс (Алгебра)
477 а) x2-3×2+2;
б) b2c2-4bc-b2-c2+1;
в) y2-10y+25-4×2;
г) (a+b)3-a3-b3;
д) x16-y16;
е) x4-3×2+1;
ж) x4-8×2+4;
з) x4-7×2+1;
и) x4+12×2+64;
к) x2+x2-2.
$$x^4-3x^2+2=x^4-2\cdot x^2\cdot \frac{3}{2}+\left(\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2+2$$
$$=\left(x^2-\frac{3}{2}\right)^2-\frac{1}{4}=\left(x^2-\frac{3}{2}-\frac{1}{2}\right)\left(x^2-\frac{3}{2}+\frac{1}{2}\right)$$
$$=(x^2-2)(x^2-1)=(x^2-2)(x-1)(x+1).$$
$$b^2c^2-4bc-b^2-c^2+1=(b^2c^2-2bc+1)-(b^2+2bc+c^2)$$
$$=(bc-1)^2-(b+c)^2=(bc-1-b-c)(bc-1+b+c).$$
$$y^2-10y+25-4x^2=(y-5)^2-(2x)^2$$
$$=(y-5-2x)(y-5+2x).$$
$$ (a+b)^3-a^3-b^3=(a+b)^3-(a^3+b^3) $$
$$=(a+b)^3-(a+b)(a^2-ab+b^2)$$
$$=(a+b)\bigl((a+b)^2-(a^2-ab+b^2)\bigr)$$
$$=(a+b)(a^2+2ab+b^2-a^2+ab-b^2)=3ab(a+b).$$
$$x^{16}-y^{16}=(x^8-y^8)(x^8+y^8)$$
$$=(x^4-y^4)(x^4+y^4)(x^8+y^8)$$
$$=(x^2-y^2)(x^2+y^2)(x^4+y^4)(x^8+y^8)$$
$$=(x-y)(x+y)(x^2+y^2)(x^4+y^4)(x^8+y^8).$$
$$x^4-3x^2+1=(x^4-2x^2+1)-x^2$$
$$=(x^2-1)^2-x^2=(x^2-1-x)(x^2-1+x).$$
$$x^4-8x^2+4=(x^4-4x^2+4)-4x^2$$
$$=(x^2-2)^2-(2x)^2=(x^2-2-2x)(x^2-2+2x).$$
$$x^4-7x^2+1=(x^4+2x^2+1)-9x^2$$
$$=(x^2+1)^2-(3x)^2=(x^2+1-3x)(x^2+1+3x).$$
$$x^4+12x^2+64=(x^4+16x^2+64)-4x^2$$
$$=(x^2+8)^2-(2x)^2=(x^2+8-2x)(x^2+8+2x).$$
$$x^4+x^2-2=\left(x^2+\frac{1}{2}\right)^2-\frac{9}{4}$$
$$=\left(x^2+\frac{1}{2}-\frac{3}{2}\right)\left(x^2+\frac{1}{2}+\frac{3}{2}\right)$$
$$=(x^2-1)(x^2+2)=(x-1)(x+1)(x^2+2).$$
Ответ
а) $$(x^2-2)(x-1)(x+1)$$
б) $$(bc-1-b-c)(bc-1+b+c)$$
в) $$(y-5-2x)(y-5+2x)$$
г) $$3ab(a+b)$$
д) $$(x-y)(x+y)(x^2+y^2)(x^4+y^4)(x^8+y^8)$$
е) $$(x^2-1-x)(x^2-1+x)$$
ж) $$(x^2-2-2x)(x^2-2+2x)$$
з) $$(x^2+1-3x)(x^2+1+3x)$$
и) $$(x^2+8-2x)(x^2+8+2x)$$
к) $$(x-1)(x+1)(x^2+2)$$
