Bai 1: thuc hien phep tinh
\(\frac{-4}{11}\).\(\frac{2}{5}\)+\(\frac{6}{11}\).\(\frac{-3}{10}\)
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a, \(\left(\frac{1}{5}-\frac{1}{4}\right)^2=\left(\frac{-1}{20}\right)^2=\frac{1}{400}\)
b, \(\left(\frac{5}{3}+\frac{1}{2}\right):\frac{-13}{5}+1\frac{5}{6}\)
=> \(\frac{13}{6}:\frac{-13}{5}+\frac{11}{6}\)
=> \(\frac{13}{6}.\frac{5}{-13}+\frac{11}{6}\)
=> \(\frac{-5}{6}+\frac{11}{6}=\frac{6}{6}=1\)
c, \(\left(\frac{3}{17}\right)^4.\left(\frac{-17}{6}\right)^4\)
=> \(\left(\frac{3}{17}.\frac{-17}{6}\right)^4=\left(\frac{-1}{2}\right)^4=\frac{1}{16}\)
-5/9+8/15+-2/11+-4/9/7/17
(-5/9+-4/9)+(8/15+7/15+-2/11
-1+1+-2/11
-2/11
(17/5+11/4)-22/5
123/20-22/5
123/20-88/20
35/20
Đặt \(A=\frac{9+\frac{9}{11}+\frac{18}{23}-\frac{27}{37}}{8+\frac{8}{11}+\frac{16}{23}-\frac{24}{37}}-\frac{2+\frac{16}{29}-\frac{24}{13}-\frac{32}{11}}{3+\frac{24}{29}-\frac{36}{13}-\frac{48}{11}}\)\(=\frac{9\left(1+\frac{1}{11}+\frac{2}{23}-\frac{3}{37}\right)}{8\left(1+\frac{1}{11}+\frac{2}{23}-\frac{3}{37}\right)}-\frac{2\left(1+\frac{8}{29}-\frac{12}{13}-\frac{16}{11}\right)}{3\left(1+\frac{8}{29}-\frac{12}{13}-\frac{16}{11}\right)}\)
\(=\frac{9}{8}-\frac{2}{3}\)(do \(1+\frac{1}{11}+\frac{2}{23}-\frac{3}{37};1+\frac{8}{29}-\frac{12}{13}-\frac{16}{11}\ne0\))
\(=\frac{27}{24}-\frac{16}{24}=\frac{11}{24}.\)
Vậy A = \(\frac{11}{24}.\)
=> \(\Rightarrow\left(\frac{11}{5}-\frac{11}{7}+\frac{11}{7}-\frac{11}{9}+...+\frac{11}{59}-\frac{11}{61}\right):2=\left(\frac{11}{5}-\frac{11}{61}\right):2=\frac{616}{305}:2=\frac{308}{305}\)
Đặt \(A=\frac{11}{5.7}+\frac{11}{7.9}+...+\frac{11}{59.61}\)
\(\Rightarrow2A:11=\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{59.61}\)
\(\Rightarrow2A:11=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{61}\)
\(\Rightarrow2A:11=\frac{1}{5}-\frac{1}{61}\)
\(\Rightarrow2A:11=\frac{56}{305}\)
\(\Rightarrow2A=\frac{56}{305}.11=\frac{616}{305}\)
\(\Rightarrow A=\frac{616}{305}:2=\frac{308}{305}\)
Vậy kết quả của phép tính trên là \(\frac{308}{305}\)
-17/55