Question

Помогите с решением 2 надо упростить выражение 4 вычитание

Answer 1

Ответ:

4) $3y-\frac{18y^{2}}{6y+1}=\frac{3y}{6y+1}$

2) $\frac{6b^{3}+48b}{b^{3}+64}-\frac{3b^{2}}{b^{2}-4b+16}=\frac{3b}{b+4}$

Объяснение:

4) $3y-\frac{18y^{2}}{6y+1}=\frac{3y(6y+1)-18y^{2}}{6y+1}=\frac{18y^{2}+3y-18y^{2}}{6y+1} =\frac{3y}{6y+1}$

2) $\frac{6b^{3}+48b}{b^{3}+64}-\frac{3b^{2}}{b^{2}-4b+16}=$

$=\frac{6b^{3}+48b}{(b+4)(b^{2}-4b+16)}-\frac{3b^{2}}{b^{2}-4b+16}=$

$=\frac{6b^{3}+48b-3b^{2}(b+4)}{(b+4)(b^{2}-4b+16)}=\frac{6b^{3}+48b-3b^{3}-12b^{2}}{(b+4)(b^{2}-4b+16)}=$

$=\frac{3b^{3}-12b^{2}+48b}{(b+4)(b^{2}-4b+16)}=\frac{3b(b^{2}-4b+16)}{(b+4)(b^{2}-4b+16)}=\frac{3b}{b+4}$