Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\frac{50-\frac{4}{13}+\frac{2}{15}-\frac{2}{17}}{100-\frac{8}{13}+\frac{4}{15}-\frac{4}{17}}\)
\(=\frac{50-\frac{4}{13}+\frac{2}{15}-\frac{2}{17}}{2\left(50-\frac{4}{13}+\frac{2}{15}-\frac{2}{17}\right)}\)
\(=\frac{1}{2}\)
\(B=\frac{1}{19}+\frac{9}{19.29}+\frac{9}{29.39}+......+\frac{9}{1999.2009}\)
\(=\frac{1}{19}+\frac{9}{10}\left(\frac{1}{19}-\frac{1}{29}+\frac{1}{19}-\frac{1}{39}+....+\frac{1}{1999}-\frac{1}{2009}\right)\)
\(=\frac{1}{19}+\frac{9}{10}\left(\frac{1}{19}-\frac{1}{2009}\right)\)
\(=\frac{1}{19}+\frac{9}{10}\cdot\frac{1990}{38171}\)
\(=\frac{200}{2009}\)
a) \(\frac{-77}{143}+\frac{65}{143}-\frac{66}{143}+\frac{7}{22}\)
= \(\frac{-78}{143}+\frac{7}{22}\)= \(\frac{-6}{11}+\frac{7}{22}\)= \(\frac{-12}{22}+\frac{7}{22}\)
= \(\frac{-5}{22}\)
b) \(\frac{-4}{5}-\frac{20}{170}+\frac{51}{170}+\frac{150}{170}\)= \(\frac{-4}{5}-\frac{221}{170}\)
\(\frac{-4}{5}-\frac{13}{10}\)= \(\frac{-8}{10}-\frac{13}{10}\)=\(\frac{-21}{10}\)
a: \(\dfrac{1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}}{1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}}:\dfrac{13+\dfrac{13}{2}+\dfrac{13}{3}+\dfrac{13}{4}}{17-\dfrac{17}{2}+\dfrac{17}{3}-\dfrac{17}{4}}\)
\(=\dfrac{1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}}{1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}}\cdot\dfrac{17\left(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\right)}{13\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}\right)}=\dfrac{17}{13}\)
b: \(\dfrac{0.125-\dfrac{1}{5}+\dfrac{1}{7}}{0.375-\dfrac{3}{5}+\dfrac{3}{7}}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-0.2}{\dfrac{3}{4}+0.5-\dfrac{3}{10}}\)
\(=\dfrac{\dfrac{1}{8}-\dfrac{1}{5}+\dfrac{1}{7}}{\dfrac{3}{8}-\dfrac{3}{5}+\dfrac{3}{7}}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}}{\dfrac{3}{4}+\dfrac{3}{6}-\dfrac{3}{10}}\)
\(=\dfrac{1}{3}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}}{\dfrac{3}{2}\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}\right)}=\dfrac{1}{3}+\dfrac{2}{3}=1\)
\(=\frac{50-\frac{4}{13}+\frac{2}{15}-\frac{2}{17}}{2\left(50-\frac{4}{13}+\frac{2}{15}-\frac{2}{17}\right)}=\frac{1}{2}\)