Упр.8.11 ГДЗ Мерзляк Поляков 7 класс (Алгебра)

1) (3a^2 b)^2;
2) (-0,2x^3 y^4 )^3;
3) (-10m^2 y^8 )^5;
4) (16x^6 y^7 z^8 )^2;
5) (-1/5 c^6 d)^4;
6) (1 1/2 a^8 b^9 )^6.

1. $$\left(3a^2b\right)^2=3^2\cdot \left(a^2\right)^2\cdot b^2=9a^4b^2.$$
2. $$\left(-0{,}2x^3y^4\right)^3=(-0{,}2)^3\cdot \left(x^3\right)^3\cdot \left(y^4\right)^3=-0{,}008x^9y^{12}.$$
3. $$\left(-10m^2y^8\right)^5=(-10)^5\cdot \left(m^2\right)^5\cdot \left(y^8\right)^5=-100000m^{10}y^{40}.$$
4. $$\left(16x^6y^7z^8\right)^2=16^2\cdot \left(x^6\right)^2\cdot \left(y^7\right)^2\cdot \left(z^8\right)^2=256x^{12}y^{14}z^{16}.$$
5. $$\left(-\frac15 c^6d\right)^4=\left(-\frac15\right)^4\cdot \left(c^6\right)^4\cdot d^4=\frac{1}{625}c^{24}d^4.$$
6. $$\left(1\frac12 a^8b^9\right)^6=\left(\frac32 a^8b^9\right)^6=\left(\frac32\right)^6\cdot \left(a^8\right)^6\cdot \left(b^9\right)^6=\frac{729}{64}a^{48}b^{54}.$$
   
   $$\frac{729}{64}=11\frac{25}{64},$$
   значит
   $$\left(1\frac12 a^8b^9\right)^6=11\frac{25}{64}a^{48}b^{54}.$$

Ответ

1) $$9a^4b^2$$; 2) $$-0{,}008x^9y^{12}$$; 3) $$-100000m^{10}y^{40}$$; 4) $$256x^{12}y^{14}z^{16}$$; 5) $$\frac{1}{625}c^{24}d^4$$; 6) $$11\frac{25}{64}a^{48}b^{54}$$.