Упр.1401 ГДЗ Мерзляк Полонский 7 класс (Алгебра)

1) 1/64*a8-b6;
2) a2b6c9+8;
3) x21y24-m12n15;
4) a6b6+1.

1. $$\frac{1}{64}a^8-b^6=\left(\frac{1}{8}a^4\right)^2-(b^3)^2$$
   $$=\left(\frac{1}{8}a^4-b^3\right)\left(\frac{1}{8}a^4+b^3\right).$$

2. $$a^3b^6c^9+8=(ab^2c^3)^3+2^3$$
   $$=(ab^2c^3+2)\left(a^2b^4c^6-2ab^2c^3+4\right).$$

3. $$x^{21}y^{24}-m^{12}n^{15}=(x^7y^8)^3-(m^4n^5)^3$$
   $$=(x^7y^8-m^4n^5)\left(x^{14}y^{16}+x^7y^8m^4n^5+m^8n^{10}\right).$$

4. $$a^6b^6+1=(a^2b^2)^3+1^3$$
   $$=(a^2b^2+1)\left(a^4b^4-a^2b^2+1\right).$$

Ответ

$$1)\ \left(\frac{1}{8}a^4-b^3\right)\left(\frac{1}{8}a^4+b^3\right);$$
$$2)\ (ab^2c^3+2)\left(a^2b^4c^6-2ab^2c^3+4\right);$$
$$3)\ (x^7y^8-m^4n^5)\left(x^{14}y^{16}+x^7y^8m^4n^5+m^8n^{10}\right);$$
$$4)\ (a^2b^2+1)\left(a^4b^4-a^2b^2+1\right).$$