Помогите, пожалуйста, упростить два выражения!!!!!! Заранее большое спасибо!!!!

1/(a+1/(b+1/c)=1/(a+c/(bc+1))=(bc+1)/(abc+a+c) 1/(a+1/b)=b/(ab+1) 1/(b(abc+a+c))-(bc+1)(ab+1)/(b(abc+a+c))=(1-(bc+1)(ab+1))/(b(abc+a+c)

[latex]\displaystyle \frac{(2p-q)^{2}+2q^{2}-3pq}{ \frac{2}{p}+q^{2} }: \frac{4p^{2}-3pq}{2+pq^{2}}= \\ \\ =\frac{4p^{2}-4pq+q^{2}+2q^{2}-3pq}{ \frac{2+pq^{2} }{p}}}* \frac{2+pq^{2}}{p(4p-3q)}= \\ \\= \frac{p(4p^{2}+3q^{2}-7pq)*(2+pq ^{2})}{p(2+pq ^{2})(4p-3q)}= \frac{4p^{2}+3q^{2}-7pq}{4p-3q} [/latex] [latex]\displaystyle \frac{1}{b(abc+a+c)}- \frac{1}{a+ \frac{1}{b+ \frac{1}{c} } }: \frac{1}{a+ \frac{1}{b}}= \\ \\ =\frac{1}{b(abc+a+c)}- \frac{1}{a+ \frac{c}{bc+1} }* \frac{ab+1}{b}= \\ \\ =\frac{1}{b(abc+a+c)}- \frac{(bc+1)(ab+1)}{b(abc+a+c)}= \frac{1-(bc+1)(ab+1)}{b(abc+a+c)}= \\ \\ = \frac{1-(ab^{2}c+ab+bc+1) }{b(abc+a+c)}=-\frac{b(abc+a+c)}{b(abc+a+c)}=-1 [/latex]