Question

Решите пожалуйста!!!!!!!!!
Даю 50 баллов ​

Answer 1

$\displaystyle\bf\\1)\\\\2x^{6} y^{-10} \cdot0,7x^{-4} y^{12} =2\cdot 0,7\cdot x^{6-4}\cdot y^{-10+12} =1,4x^{2}y^{2} \\\\\\2)\\\\3,6a^{-8} c^{4} \cdot(-3a^{-3}c)^{-2} =3,6a^{-8} c^{4} \cdot\Big(\frac{1}{9} a^{6} c^{-2} \Big)=\\\\=3,6\cdot\frac{1}{9} a^{-8+6} c^{4-2} =0,4a^{-2} c^{2} \\\\\\3)\\\\\Big(\frac{3x^{-2} }{4y^{-3} } \Big)^{-1} \cdot6xy^{4} =\Big(\frac{3y^{3} }{4x^{2} } \Big)^{-1} \cdot6xy^{4} =\frac{4x^{2}\cdot 6xy^{4} }{3y^{3} } =\\\\=\frac{8x^{3} y^{4} }{y^{3} } =8x^{3} y$

Answer 2

$1)2x^6y^-^1^0*0,7x^-^4y^1^2=1,4x^6^-^4y^-^1^0^+^1^2=1,4x^2y^2\\\\ 2)3,6a^-^8c^4*(-3a^-^3c)^-^2=\displaystyle\frac{18}{5}a^-^8c^4*(3a^-^3c)^-^2=\frac{18}{5}a^-^8c^4*3^-^2a^6c^-^2=\\ \\ \frac{18}{5}a^-^8c^4*\frac{1}{3^2}*a^6c^-^2=\frac{18}{5}a^-^8c^4*\frac{1}{9}a^6c^-^2=\frac{2}{5}a^-^8c^4a^6c^-^2=\frac{2}{5}a^-^2c^2=\frac{2}{5}*\\ \\ \frac{1}{a^2}*c^2=\frac{2c^2}{5a^2}\\\\ 3)(\frac{3x^-^2}{4y^-^3})^-^1*6xy^4=\frac{4y^-^3}{3x^-^2}*6xy^4=4y^-^3*2x^3y^4=8x^3y$