\begin{gathered}1) (4m+ \frac{1}{3}n )^{3}=64m^{3}+ 16m^{2}n+ \frac{4}{3} mn^{2}+\frac{1}{27}n^{3} \\ 2) ( \frac{2}{3}x-3y )^{3}= \frac{8}{27} x^{3}-4x^{2}y+18xy^{2}-27y^{3} \\ 3) ( \frac{1}{3} a+ \frac{1}{2} b)^{3}= \frac{1}{27} a^{3}+ \frac{1}{6} a^{2}b+ \frac{1}{4} ab^{2}+ \frac{1}{8} b^{3} \\ 4) ( \frac{1}{6} x+2y)}^{3}= \frac{1}{216} x^{3}+ \frac{1}{6} x^{2}y+2xy^{2}+8y^{3} \\ 5) (0,2x-5y)^{3}=0,008x^{3}-0,6x^{2}y+15xy^{2}-125y^{3} \\ 6) (3a-0,6b)^{3}=27a^{3}-16,2a^{2}b+3,24ab^{2}-0,216b^{3} \\end{gathered} \begin{gathered}7) (0,1m-4n)^{3}= 0,001m^{3}-0,12m^{2}n+4,8mn^{2}-64n^{3} \\ 8) (0,5a+0,16)^{3}= 0,125a^{3}+0,12a^{2}+0,0384a+0,004096\end{gathered} 7)(0,1m−4n)
3
=0,001m
3
−0,12m
2
n+4,8mn
2
−64n
3

8)(0,5a+0,16)
3
=0,125a
3
+0,12a
2
+0,0384a+0,004096
​ ​

potehin30    3   25.01.2021 15:17    0