Упр.216 ГДЗ Никольский Потапов 7 класс (Алгебра)
б) (-1*2/3)b2c3 * (-2/15)*b2c2;
в) 1.2*ck2*2/3*ck;
г) 1*2/3*k3p2*(-1*1/5)kp2;
д) (-2*1/4)p2x2*1*1/3*px3;
е) (-9/11)*x2y3 * (-1*2/9)xy;
ж) (-1*2/3)a2x3*(-3/5)a2x4;
з) (-285/6)a3c2*1*2/3*ac2.
а) $$1\frac{1}{5}a^2b^3\cdot 1\frac{1}{9}ab^2=\frac{6}{5}a^2b^3\cdot \frac{10}{9}ab^2=\frac{60}{45}a^3b^5=\frac{4}{3}a^3b^5=1\frac{1}{3}a^3b^5.$$
б) $$\left(-1\frac{2}{3}\right)b^2c^3\cdot \left(-\frac{2}{15}\right)b^2c^2=-\frac{5}{3}b^2c^3\cdot \left(-\frac{2}{15}\right)b^2c^2=\frac{10}{45}b^4c^5=\frac{2}{9}b^4c^5.$$
в) $$1\frac{1}{2}ck^2\cdot \frac{2}{3}ck=\frac{3}{2}ck^2\cdot \frac{2}{3}ck=c^2k^3.$$
г) $$1\frac{2}{3}k^3p^2\cdot \left(-1\frac{1}{5}\right)kp^2=\frac{5}{3}k^3p^2\cdot \left(-\frac{6}{5}\right)kp^2=-2k^4p^4.$$
д) $$\left(-2\frac{1}{4}\right)p^2x^2\cdot 1\frac{1}{3}px^3=-\frac{9}{4}p^2x^2\cdot \frac{4}{3}px^3=-3p^3x^5.$$
е) $$\left(-\frac{9}{11}\right)x^2y^3\cdot \left(-1\frac{2}{9}\right)xy=\left(-\frac{9}{11}\right)x^2y^3\cdot \left(-\frac{11}{9}\right)xy=x^3y^4.$$
ж) $$\left(-1\frac{2}{3}\right)a^2x^3\cdot \left(-\frac{3}{5}\right)a^2x^4=\left(-\frac{5}{3}\right)a^2x^3\cdot \left(-\frac{3}{5}\right)a^2x^4=a^4x^7.$$
з) $$\left(-2\frac{5}{6}\right)a^3c^2\cdot 1\frac{2}{3}ac^2=\left(-\frac{17}{6}\right)a^3c^2\cdot \frac{5}{3}ac^2=-\frac{85}{18}a^4c^4=-4\frac{13}{18}a^4c^4.$$
Ответ
а) $$1\frac{1}{3}a^3b^5$$; б) $$\frac{2}{9}b^4c^5$$; в) $$c^2k^3$$; г) $$-2k^4p^4$$; д) $$-3p^3x^5$$; е) $$x^3y^4$$; ж) $$a^4x^7$$; з) $$-4\frac{13}{18}a^4c^4$$.
