Упр.270 ГДЗ Муравин 7 класс (Алгебра)

1) (-2ах^2 у^3)^4; 4) (-4а^2 b^5) · (0,5аb^3)^3;
2) (-0,3b^5 cx^3)^9; 5) (2 1/3 x^2 y^3)^4 · (3/7 x^5 y^4)^4;
3) (-3b^4 y)^2 · 5by^8; 6) (-3 1/3 k^3 p^4)^3 · (-0,3k^2 p)^3.

1. $$(-2ax^2y^3)^4=(-2)^4a^4(x^2)^4(y^3)^4=16a^4x^8y^{12}$$
2. $$(-0{,}3b^5cx^3)^9=(-0{,}3)^9(b^5)^9c^9(x^3)^9=-0{,}3^9b^{45}c^9x^{27}$$
3. $$(-3b^4y)^2\cdot 5by^8=(-3)^2(b^4)^2y^2\cdot 5by^8=9b^8y^2\cdot 5by^8=45b^9y^{10}$$
4. $$(-4a^2b^5)\cdot(0{,}5ab^3)^3=-4a^2b^5\cdot(0{,}5)^3a^3(b^3)^3$$
   $$=-4a^2b^5\cdot 0{,}125a^3b^9=-0{,}5a^5b^{14}$$
5. $$\left(2\frac{1}{3}x^2y^3\right)^4\cdot\left(\frac{3}{7}x^5y^4\right)^4 =\left(\frac{7}{3}\right)^4(x^2)^4(y^3)^4\cdot\left(\frac{3}{7}\right)^4(x^5)^4(y^4)^4$$
   $$=\left(\frac{7}{3}\cdot\frac{3}{7}\right)^4x^{8+20}y^{12+16}=x^{28}y^{28}$$
6. $$\left(-3\frac{1}{3}k^3p^4\right)^3\cdot(-0{,}3k^2p)^3 =\left(-\frac{10}{3}\right)^3(k^3)^3(p^4)^3\cdot\left(-\frac{3}{10}\right)^3(k^2)^3p^3$$
   $$=\left(-\frac{10}{3}\cdot-\frac{3}{10}\right)^3k^{9+6}p^{12+3}=k^{15}p^{15}$$

Ответ

1) $$16a^4x^8y^{12}$$; 2) $$-0{,}3^9b^{45}c^9x^{27}$$; 3) $$45b^9y^{10}$$; 4) $$-0{,}5a^5b^{14}$$; 5) $$x^{28}y^{28}$$; 6) $$k^{15}p^{15}$$.