Задание 547

Решение:

• \(\begin{array}[t]{ll}{\largeПри}\ a=4,\ b=2&&\phantom{0000}\frac{a}{b+a}=\frac{4}{2+4}=\frac{4}{6}=\frac{2}{3}\end{array}\)

• \(\begin{array}[t]{ll}{\largeПри}\ a={-}10,\ b=20&&\phantom{\ }\frac{a}{b+a}=\frac{{-}10}{20+({-}10)}=\frac{{-}10}{20-10}=\frac{{-}10}{10}={-}1\end{array}\)

• \(\begin{array}[t]{ll}{\largeПри}\ a=6,\ b={-}5&&\phantom{\ 00}\frac{a}{b+a}=\frac{6}{{-}5+6}=\frac{6}{1}=6\end{array}\)

• \(\begin{array}[t]{ll}{\largeПри}\ a=0,\ b=7&&\phantom{\ 0000}\frac{a}{b+a}=\frac{0}{7+0}=\frac{0}{7}=0\end{array}\)

• \(\begin{array}[t]{ll}{\largeПри}\ a={-1},\ b=0&&\phantom{000}\frac{a}{b+a}=\frac{{-}1}{0+({-}1)}=\frac{{-1}}{0-1}=\frac{{-}1}{{-}1}=1\end{array}\)

• \(\begin{array}[t]{ll}{\largeПри}\ a={-}\frac{1}{2},\ b={-}2&&\phantom{\ }\frac{a}{b+a}=\frac{{-}\frac{1}{2}}{{-}2+\left({-}\frac{1}{2}\right)}=\frac{{-}\frac{1}{2}}{{-}2\frac{1}{2}}=\frac{{-}\frac{1}{2}}{{-}\frac{5}{2}}={-}\frac{1}{2}:\left({-}\frac{5}{2}\right)={-}\frac{1}{2}\cdot\left({-}\frac{2}{5}\right)=\frac{1}{5}\end{array}\)

• \(\begin{array}[t]{ll}{\largeПри}\ a={-}0{,}7,\ b=1{,}4&&\frac{a}{b+a}=\frac{{-}0{,}7}{1{,}4+({-}0{,}7)}=\frac{{-}0{,}7}{1{,}4-0{,}7}=\frac{{-}0{,}7}{0{,}7}={-}1\end{array}\)