Задание 552

Решение:

• \(\frac{a}{b}\)

• \(\begin{array}[t]{ll}{\largeПри}\ a=2,\ b=1&&\frac{a}{b}=\frac{2}{1}=2\\\\{\largeПри}\ a={-}1,\ b={-}3&&\frac{a}{b}=\frac{{-}1}{{-3}}=\frac{1}{3}\\\\{\largeПри}\ a=\frac{1}{2},\ b=0{,}2&&\frac{a}{b}=\frac{\frac{1}{2}}{0{,}2}=\frac{\frac{1}{2}}{\frac{1}{5}}=\frac{1}{2}:\frac{1}{5}=\frac{1}{2}\cdot\frac{5}{1}=\frac{5}{2}=2\frac{1}{2}=2{,}5\\\\{\largeПри}\ a=0{,}4,\ b={-}\frac{1}{3}&&\frac{a}{b}=\frac{0{,}4}{{-}\frac{1}{3}}=\frac{\frac{2}{5}}{{-}\frac{1}{3}}=\frac{2}{5}:\left({-}\frac{1}{3}\right)=\frac{2}{5}\cdot\left({-}\frac{3}{1}\right)={-}\frac{6}{5}={-}1\frac{1}{5}={-}1{,}2\end{array}\)

• \(a-\frac{1}{b}\)

• \(\begin{array}[t]{ll}{\largeПри}\ a=2,\ b=1&&a-\frac{1}{b}=2-\frac{1}{1}=2-1=1\\\\{\largeПри}\ a={-}1,\ b={-}3&&a-\frac{1}{b}={-}1-\frac{1}{{-3}}={-}1+\frac{1}{3}={-}\frac{3}{3}+\frac{1}{3}={-}\frac{2}{3}\\\\{\largeПри}\ a=\frac{1}{2},\ b=0{,}2&&a-\frac{1}{b}=\frac{1}{2}-\frac{1}{0{,}2}=\frac{1}{2}-5=\frac{1}{2}-\frac{10}{2}={-}\frac{9}{2}={-}4\frac{1}{2}={-}4{,}5\\\\{\largeПри}\ a=0{,}4,\ b={-}\frac{1}{3}&&a-\frac{1}{b}=0{,}4-\frac{1}{{-}\frac{1}{3}}=0{,}4-\frac{1}{1}:\left({-}\frac{1}{3}\right)=0{,}4-\frac{1}{1}\cdot\left({-}\frac{3}{1}\right)=0{,}4+3=3{,}4\end{array}\)

• \(\frac{a+b}{a}\)

• \(\begin{array}[t]{ll}{\largeПри}\ a=2,\ b=1&&\frac{a+b}{a}=\frac{2+1}{2}=\frac{3}{2}=1\frac{1}{2}=1{,}5\\\\{\largeПри}\ a={-}1,\ b={-}3&&\frac{a+b}{a}=\frac{{-}1+({-}3)}{{-}1}=\frac{{-}4}{{-}1}=4\\\\{\largeПри}\ a=\frac{1}{2},\ b=0{,}2&&\frac{a+b}{a}=\frac{\frac{1}{2}+0{,}2}{\frac{1}{2}}=\frac{\frac{1}{2}+\frac{1}{5}}{\frac{1}{2}}=\frac{\frac{5+2}{10}}{\frac{1}{2}}=\frac{7}{10}:\frac{1}{2}=\frac{7}{10}\cdot\frac{2}{1}=\frac{7}{5}\cdot\frac{1}{1}=\frac{7}{5}=1\frac{2}{5}=1{,}4\\\\{\largeПри}\ a=0{,}4,\ b={-}\frac{1}{3}&&\frac{a+b}{a}=\frac{0{,}4+\left({-}\frac{1}{3}\right)}{0{,}4}=\frac{\frac{2}{5}-\frac{1}{3}}{\frac{2}{5}}=\frac{\frac{6-5}{15}}{\frac{2}{5}}=\frac{1}{15}:\frac{2}{5}=\frac{1}{15}\cdot\frac{5}{2}=\frac{1}{3}\cdot\frac{1}{2}=\frac{1}{6}\end{array}\)

• \(\frac{a-b}{a+b}\)

• \(\begin{array}[t]{ll}{\largeПри}\ a=2,\ b=1&&\frac{a-b}{a+b}=\frac{2-1}{2+1}=\frac{1}{3}\\\\{\largeПри}\ a={-}1,\ b={-}3&&\frac{a-b}{a+b}=\frac{{-}1-({-}3)}{{-}1+({-}3)}=\frac{{-}1+3}{{-}4}=\frac{2}{{-}4}={-}\frac{1}{2}={-}0{,}5\\\\{\largeПри}\ a=\frac{1}{2},\ b=0{,}2&&\frac{a-b}{a+b}=\frac{\frac{1}{2}-0{,}2}{\frac{1}{2}+0{,}2}=\frac{\frac{1}{2}-\frac{1}{5}}{\frac{1}{2}+\frac{1}{5}}=\frac{\frac{5-2}{10}}{\frac{5+2}{10}}=\frac{\frac{3}{10}}{\frac{7}{10}}=\frac{3}{10}:\frac{7}{10}=\frac{3}{10}\cdot\frac{10}{7}=\frac{3}{1}\cdot\frac{1}{7}=\frac{3}{7}\\\\{\largeПри}\ a=0{,}4,\ b={-}\frac{1}{3}&&\frac{a-b}{a+b}=\frac{0{,}4-\left({-}\frac{1}{3}\right)}{0{,}4+\left({-}\frac{1}{3}\right)}=\frac{\frac{2}{5}+\frac{1}{3}}{\frac{2}{5}-\frac{1}{3}}=\frac{\frac{6+5}{15}}{\frac{6-5}{15}}=\frac{\frac{11}{15}}{\frac{1}{15}}=\frac{11}{15}:\frac{1}{15}=\frac{11}{15}\cdot\frac{15}{1}=\frac{11}{1}\cdot\frac{1}{1}=\frac{11}{1}=11\end{array}\)

• \(\frac{a^2-b^2}{a-2b}\)

• \(\begin{array}[t]{ll}{\largeПри}\ a=2,\ b=1&&\frac{a^2-b^2}{a-2b}=\frac{2^2-1^2}{2-2\cdot1}=\frac{4-1}{2-2}=\frac{3}{0}\ -\ {\largeвыражение\ не\ имеет\ смысла}\\\\{\largeПри}\ a={-}1,\ b={-}3&&\frac{a^2-b^2}{a-2b}=\frac{ ({-}1)^2-({-}3)^2}{{-}1-2\cdot({-}3)}=\frac{1-9}{{-}1+6}=\frac{{-}8}{5}={-}1\frac{3}{5}={-}1{,}6\\\\{\largeПри}\ a=\frac{1}{2},\ b=0{,}2&&\frac{a^2-b^2}{a-2b}=\frac{\left(\frac{1}{2}\right)^2-0{,}2^2}{\frac{1}{2}-2\cdot0{,}2}=\frac{\frac{1}{4}-0{,}04}{0{,}5-0{,}4}=\frac{0{,}25-0{,}04}{0{,}1}=\frac{0{,}21}{0{,}1}=2{,}1\\\\{\largeПри}\ a=0{,}4,\ b={-}\frac{1}{3}&&\frac{a^2-b^2}{a-2b}=\frac{0{,}4^2-\left({-}\frac{1}{3}\right)^2}{0{,}4-2\cdot\left({-}\frac{1}{3}\right)}=\frac{0{,}16-\frac{1}{9}}{0{,}4+\frac{2}{3}}=\frac{\frac{4}{25}-\frac{1}{9}}{\frac{2}{5}+\frac{2}{3}}=\frac{\frac{36-25}{225}}{\frac{6+10}{15}}=\frac{\frac{11}{225}}{\frac{16}{15}}=\frac{11}{225}:\frac{16}{15}=\frac{11}{225}\cdot\frac{15}{16}=\frac{11}{15}\cdot\frac{1}{16}=\frac{11}{240}\end{array}\)