Question

Решите плизз дам максимум баллов 6 задание ​

Answer 1

$1)-3\frac{1}{5}a^{8}b(\frac{1}{2}a^{3}b^{8})^{4}= -\frac{16}{5}a^{8}b*\frac{1}{16}a^{12}b^{32}=-\frac{1}{5}a^{8+12}bx^{1+32}=-0,2a^{20}b^{33}\\\\2)x^{n-2}*x^{2}*x^{n+2}=x^{n-2+2+n+2}=x^{2n+2}$

Answer 2

4 (a)

$-3a^{5} *4ab^{6}=-12a^{5+1}b^{6}=-12a^{6}b^{6}$

4 (б)

$(-2xy^{6})^{4}=16x^{4}y^{24}$

4 (в)

$(-3a^{3}b^{4})^3=(-3)^3a^{3*3}b^{4*3}=-27a^9b^{12}$

5 (a)

$\frac{6^{15}*6^{11}}{6^{24}}=6^{15+11-24}=6^{2}=36$

5 (б)

$\frac{3^{11}*27}{9^{6}}=\frac{3^{11}*3^{3}}{(3^2)^{6}}=3^{11+3-12}=3^{2}=9$

6(a)

$-3\frac{1}{5}a^{8}b*(\frac{1}{2}a^{3}b^{8})^4=-\frac{16}{5}a^{8}b*(\frac{1}{2})^4a^{3*4}b^{8*4}=$

$=-\frac{16}{5}*\frac{1}{16}a^{8+12}b^{1+32}=-\frac{1}{5}a^{20}b^{33}$

6 (б)

$x^{n-2}*x^{2}*x^{n+2}=x^{n-2+2+n+2}=x^{2n+2}$