Найдите ab/c, если ab/(a+b)=1; ac/(a+c)=2; bc/(b+c)=3.

$\frac{ab}{a+b}=1\\ \frac{1}{\frac{a+b}{ab}}=1 \\ \frac{a+b}{ab}=1\\ \frac{1}{a}+\frac{1}{b}=1\\ \\ \frac{ac}{a+c}=2\\ \frac{1}{\frac{a+c}{ac}}=2\\ \frac{a+c}{ac}=\frac{1}{2}\\ \frac{1}{a}+\frac{1}{c} \\\\ \frac{bc}{b+c}=3 \\ \frac{1}{b}+\frac{1}{c}=\frac{1}{3}\\ \\$
$\frac{1}{a}+\frac{1}{b}=1\\ \frac{1}{a}+\frac{1}{c}=\frac{1}{2}\\ \frac{1}{b}+\frac{1}{c }=\frac{1}{3}\\\\ \frac{1}{b}-\frac{1}{c}=\frac{1}{2}\\ \frac{1}{b}+\frac{1}{c}=\frac{1}{3}\\\\ \frac{2}{b}=\frac{5}{6}\\ b=\frac{12}{5}\\ c=-12 \\ a=\frac{12}{7}\\ \frac{ab}{c}=\frac{-12}{35}$