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\(A=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{48.50}.\)
\(=\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}....+\frac{2}{48.50}\right)\)
\(=\frac{1}{2}.\left(\frac{4-2}{2.4}+\frac{6-4}{4.6}+\frac{8-6}{6.8}+...+\frac{50-48}{48.50}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+.....+\frac{1}{48}-\frac{1}{50}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{50}\right)\)
\(=\frac{1}{2}.\frac{12}{25}=\frac{6}{25}\)
\(B=\frac{3}{1.4}+\frac{3}{4.7}+....+\frac{3}{97.100}\)
\(=\frac{4-1}{1.4}+\frac{7-4}{4.7}+....+\frac{100-97}{97.100}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{97}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
\(C=\frac{8}{7.14}+\frac{8}{14.21}+....+\frac{8}{91.98}\)
\(=\frac{7}{8}.\left(\frac{7}{7.14}+\frac{7}{14.21}+...+\frac{7}{91.98}\right)\)
\(=\frac{7}{8}.\left(\frac{1}{7}-\frac{1}{14}+\frac{1}{14}-\frac{1}{21}+.....+\frac{1}{91}-\frac{1}{98}\right)\)
\(=\frac{7}{8}.\left(\frac{1}{7}-\frac{1}{98}\right)\)
\(=\frac{7}{8}.\frac{13}{98}=\frac{13}{112}\)
Nhiều thế :( Làm 1,2 câu thôi nhé
a) \(\frac{1}{3}+\frac{1}{4}=\frac{4}{12}+\frac{3}{12}=\frac{7}{12}\) (bị mất nét nhưng vẫn nhìn ra là số 12 nhỉ?)
b) \(\frac{-2}{5}+\frac{7}{21}=\frac{-42}{105}+\frac{35}{105}=\frac{-7}{105}=\frac{-1}{15}\)
Bài 1:
a) Ta có: \(\frac{3}{8}+\frac{-5}{6}\)
\(=\frac{3}{8}-\frac{5}{6}\)
\(=\frac{9}{24}-\frac{20}{24}\)
\(=-\frac{11}{24}\)
b) Ta có: \(\frac{15}{12}-\frac{-1}{4}\)
\(=\frac{15}{12}+\frac{1}{4}\)
\(=\frac{15}{12}+\frac{3}{12}\)
\(=\frac{18}{12}=\frac{3}{2}\)
Bài 2:
a) Ta có: \(-\frac{1}{12}-\left(2\frac{5}{8}-\frac{1}{3}\right)\)
\(=-\frac{1}{12}-\frac{21}{8}+\frac{1}{3}\)
\(=\frac{-2}{24}-\frac{63}{24}+\frac{8}{24}\)
\(=\frac{-57}{24}\)
\(=-\frac{19}{8}\)
b) Ta có: \(\frac{-5}{6}-\left(\frac{-3}{8}+\frac{1}{10}\right)\)
\(=\frac{-5}{6}+\frac{3}{8}-\frac{1}{10}\)
\(=\frac{-100}{120}+\frac{45}{120}-\frac{12}{120}\)
\(=\frac{-67}{120}\)
c) Ta có: \(-1.75-\left(\frac{-1}{9}-2\frac{1}{18}\right)\)
\(=-\frac{7}{4}+\frac{1}{9}+\frac{37}{18}\)
\(=\frac{-63}{36}+\frac{4}{36}+\frac{74}{36}\)
\(=\frac{15}{36}=\frac{5}{12}\)
#)Giải :
a) x + 2x + 3x + ... + 100x = - 213
=> 100x + ( 2 + 3 + 4 + ... + 100 ) = - 213
=> 100x + 5049 = - 213
<=> 100x = - 5262
<=> x = - 52,62
#)Giải :
b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}x-\frac{1}{6}\)
\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{3}+\frac{1}{6}\)
\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{2}\)
\(\Rightarrow\left(\frac{1}{2}+\frac{1}{4}\right)x=\frac{1}{2}\)
\(\Rightarrow\frac{3}{4}x=\frac{1}{2}\)
\(\Leftrightarrow x=\frac{2}{3}\)
1/ (69.210+1210)+(219.273+15.49.94) = 29.39.210+310.220+219.39+5.3.218.38 = 219.39+310.220+219.39+5.218.39
= 218.39(2+3.22+5)=19.218.39
Ta có :
\(C=\frac{2}{1.2}+\frac{4}{2.4}+\frac{6}{4.7}+\frac{8}{7.11}+\frac{10}{11.16}+\frac{12}{16.22}\)
\(C=2\left(\frac{1}{1.2}+\frac{2}{2.4}+\frac{3}{4.7}+\frac{4}{7.11}+\frac{5}{11.16}+\frac{6}{16.22}\right)\)
\(C=2\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{22}\right)\)
\(C=2\left(1-\frac{1}{22}\right)\)
\(C=2-\frac{1}{11}\)
\(C=\frac{21}{11}\)
Vậy \(C=\frac{21}{11}\)
Chúc bạn học tốt ~