Question

Помогите пожалуйста!

$a) \frac{1}{b^3} - \frac{2}{b^2} + \frac{1}{b} b) \frac{x+y}{xy} - \frac{x+z}{xz} + \frac{y+z}{yz} c) \frac{3x+1}{3x} - \frac{2y+1}{2y} + \frac{3x-y}{6xy}$

Answer 1

Ответ: ...........

Объяснение:

Answer 2

$\tt\displaystyle a)~\frac{1}{b^3}-\frac{2}{b^2}+\frac{1}{b}=\frac{1-2b+b^2}{b^3} =\\\\\\=\frac{b^2-2b+1}{b^3}=\frac{(b-1)^2}{b^3};\\\\\\b)~\frac{x+y}{xy} -\frac{x+z}{xz}+\frac{y+z}{yz}=\\\\\\=\frac{z(x+y)-y(x+z)+x(y+z)}{xyz}=\\\\\\=\frac{xz+yz-xy-yz+xy+xz}{xyz}=\\\\\\=\frac{2xz}{xyz}=\frac{2}{y};\\\\\\c)~\frac{3x+1}{3x} -\frac{2y+1}{2y}+\frac{3x-y}{6xy}=\\\\\\=\frac{2y(3x+1)-3x(2y+1)+3x-y}{6xy}=\\\\\\=\frac{6xy+2y-6xy-3x+3x-y}{6xy}=\\\\\\=\frac{y}{6xy}=\frac{1}{6x}$