Question

Помогите номер 4 и 6 пожалуйста!!!!

Answer 1

4.

$\displaystyle x^3*\sqrt{64x^2}=x^3*|8x|=\bigg[x\geq 0\bigg]= x^3*8x=8x^4\\\\-7y^2*\sqrt{\frac{9}{y^2}}=-7y^2*|\frac{3}{y}|=\bigg[y<0\bigg]=-7y^2*\frac{3}{-y}=21y$

6.

$\displaystyle\frac{x-3}{16-\sqrt{x} } \\\\\left \{ {{x\geq 0} \atop {16-\sqrt{x} \neq 0}} \right. ;\left \{ {{x\geq 0} \atop {\sqrt{x} \neq 16}} \right.;\left \{ {{x\geq 0} \atop {x\neq 256}} \right. \\\\x \in [0;256) (256;+oo)$