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\(A=\frac{5}{13}+\frac{-5}{7}+\frac{-20}{41}+\frac{8}{13}+\frac{-21}{41}\)
\(\Leftrightarrow A=\left(\frac{5}{13}+\frac{8}{13}\right)+\left(\frac{-20}{41}+\frac{-21}{41}\right)+\frac{-5}{7}\)
\(\Leftrightarrow A=1+\left(-1\right)+\frac{-5}{7}\)
\(\Leftrightarrow A=0+\frac{-5}{7}=\frac{-5}{7}\)
Vậy A = \(\frac{-5}{7}\)
B= \(\frac{-5}{9}+\frac{8}{15}+\frac{-2}{11}+\frac{4}{-9}+\frac{7}{15}\)
\(\Leftrightarrow B=\frac{-5}{9}+\frac{8}{15}+\frac{-2}{11}+\frac{-4}{9}+\frac{7}{15}\)
\(\Leftrightarrow B=\left(\frac{-5}{9}+\frac{-4}{9}\right)+\left(\frac{8}{15}+\frac{7}{15}\right)+\frac{-2}{11}\)
\(\Leftrightarrow B=-1+1+\frac{-2}{11}\)
\(\Leftrightarrow B=0+\frac{-2}{11}\)
\(\Leftrightarrow\) \(B=\frac{-2}{11}\)
Vậy \(B=\frac{-2}{11}\)
@@ Học tốt
Chiyuki Fujito
K cần tk nhá
a, \(\frac{3}{8}+\frac{11}{13}-\frac{9}{13}\)
=\(\frac{3}{8}+\frac{2}{13}\)
=\(\frac{55}{104}.\)
b, \(\frac{2}{7}.\left(\frac{5}{9}+\frac{4}{9}\right)+\frac{2}{7}\)
=\(\frac{2}{7}.\frac{9}{9}+\frac{2}{7}\)
=\(\frac{2}{7}+\frac{2}{7}\)
=\(\frac{4}{7}\)
c, \(\frac{3}{11}.\left(\frac{3}{5}-\frac{5}{3}\right)-\frac{3}{10}.\left(\frac{1}{3}-\frac{2}{5}\right)\)
=\(\frac{3}{11}.-\frac{16}{15}-\frac{3}{10}.-\frac{1}{15}\)
=\(-\frac{16}{55}--\frac{1}{50}\)
=\(-\frac{149}{550}.\)
d, \(\frac{-3}{4}.\frac{11}{23}+\frac{3}{23}.\frac{31}{17}-\frac{3}{17}.\frac{19}{23}\)
=\(-\frac{33}{92}+\frac{93}{391}-\frac{57}{391}\)
=\(-\frac{417}{1564}\)
e, \(\frac{3}{17}.\frac{11}{23}+\frac{3}{23}.\frac{31}{17}-\frac{3}{17}.\frac{19}{23}\)
=\(\frac{33}{391}+\frac{93}{391}--\frac{254}{391}\)
=\(\frac{380}{391}.\)
g, \(\frac{3}{7}.\frac{-5}{12}+\frac{11}{17}:\frac{5}{-12}\)
=\(-\frac{5}{28}+-\frac{132}{85}\)
= \(-1.731512605.\)
k cho mình nha làm mỏi tay quá ,.....................kết bạn với mình nha.......................
a) \(\frac{4}{11}-\frac{7}{15}+\frac{7}{11}-\frac{5}{15}\)
\(=\left(\frac{4}{11}+\frac{7}{11}\right)-\left(\frac{7}{15}+\frac{5}{15}\right)\)
\(=1-\frac{4}{5}\)
\(=\frac{1}{5}\)
b) \(\frac{7}{3}-\frac{4}{9}-\frac{1}{3}-\frac{5}{9}\)
\(=\left(\frac{7}{3}-\frac{1}{3}\right)-\left(\frac{4}{9}+\frac{5}{9}\right)\)
\(=2-1\)
\(=1\)
c) \(\frac{1}{4}+\frac{7}{33}-\frac{5}{3}\)
\(=\frac{-1}{4}+\frac{-16}{11}\)
\(=\frac{-75}{44}\)
d) \(\frac{-3}{4}\times\frac{8}{11}-\frac{3}{11}\times\frac{1}{2}\)
\(=\frac{-6}{11}-\frac{3}{22}\)
\(=\frac{15}{22}\)
e) \(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\)
\(=\frac{1}{3\times5}+\frac{1}{5\times7}+\frac{1}{7\times9}+\frac{1}{9\times11}+\frac{1}{11\times13}+\frac{1}{13\times15}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\)
\(=\frac{1}{3}-\frac{1}{15}\)
\(=\frac{4}{15}\)
\(T=\left(\frac{1}{2}+1\right).\left(\frac{1}{3}+1\right).\left(\frac{1}{4}+1\right).......\left(\frac{1}{98}+1\right).\left(\frac{1}{99}+1\right)\)
\(T=\left(\frac{1}{2}+\frac{2}{2}\right).\left(\frac{1}{3}+\frac{3}{3}\right).\left(\frac{1}{4}+\frac{4}{4}\right).....\left(\frac{1}{98}+\frac{98}{98}\right).\left(\frac{1}{99}+\frac{99}{99}\right)\)
\(T=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{99}{98}.\frac{100}{99}\)
\(T=\frac{3.4.5....99.100}{2.3.4.....98.99}\)
\(T=\frac{100}{2}\)
\(T=50\)
Vậy T = 50
Chúc bạn học tốt!
\(a.\frac{108}{119}.\frac{107}{211}+\frac{108}{119}.\frac{104}{211}=\frac{108}{119}.\left(\frac{107}{211}+\frac{104}{211}\right)=\frac{108}{119}.1=108\)