Упр.1003 ГДЗ Макарычев 7 класс (Алгебра)
a)27/64-y12;
б)-х15 + 1/27;
в) 3*3a15/8 +b12;
г)1*61×18/64 +у3.
Используем формулы суммы и разности кубов:
$$a^3-b^3=(a-b)(a^2+ab+b^2),$$
$$a^3+b^3=(a+b)(a^2-ab+b^2).$$
$$\frac{27}{64}-y^{12}=\left(\frac{3}{4}\right)^3-(y^4)^3$$
$$\frac{27}{64}-y^{12}=\left(\frac{3}{4}-y^4\right)\left(\frac{9}{16}+\frac{3}{4}y^4+y^8\right)$$
$$-x^{15}+\frac{1}{27}=\left(\frac{1}{3}\right)^3-(x^5)^3$$
$$-x^{15}+\frac{1}{27}=\left(\frac{1}{3}-x^5\right)\left(\frac{1}{9}+\frac{1}{3}x^5+x^{10}\right)$$
$$3\frac{3}{8}a^{15}+b^{12}=\frac{27}{8}a^{15}+b^{12}=\left(\frac{3}{2}a^5\right)^3+(b^4)^3$$
$$3\frac{3}{8}a^{15}+b^{12}=\left(\frac{3}{2}a^5+b^4\right)\left(\frac{9}{4}a^{10}-\frac{3}{2}a^5b^4+b^8\right)$$
$$1\frac{61}{64}x^{18}+y^3=\frac{125}{64}x^{18}+y^3=\left(\frac{5}{4}x^6\right)^3+y^3$$
$$1\frac{61}{64}x^{18}+y^3=\left(\frac{5}{4}x^6+y\right)\left(\frac{25}{16}x^{12}-\frac{5}{4}x^6y+y^2\right)$$
Ответ
a) $$\left(\frac{3}{4}-y^4\right)\left(\frac{9}{16}+\frac{3}{4}y^4+y^8\right)$$
б) $$\left(\frac{1}{3}-x^5\right)\left(\frac{1}{9}+\frac{1}{3}x^5+x^{10}\right)$$
в) $$\left(\frac{3}{2}a^5+b^4\right)\left(\frac{9}{4}a^{10}-\frac{3}{2}a^5b^4+b^8\right)$$
г) $$\left(\frac{5}{4}x^6+y\right)\left(\frac{25}{16}x^{12}-\frac{5}{4}x^6y+y^2\right)$$
