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sửa đề câu a và câu b nhá , mik nghĩ đề như này :
\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{213\cdot215}\)
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{213}-\frac{1}{215}\)
= \(\frac{1}{1}-\frac{1}{215}\)
\(=\frac{214}{215}\)
b, đặt \(A=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{213\cdot215}\)
\(A\cdot2=\frac{2}{1\cdot3}+\frac{2}{3.5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{213\cdot215}\)
\(A\cdot2=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{213}-\frac{1}{215}\)
\(A\cdot2=\frac{1}{1}-\frac{1}{215}\)
\(A\cdot2=\frac{214}{215}\)
\(A=\frac{214}{215}:2\)
\(A=\frac{107}{215}\)
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Giải:
\(B=\dfrac{3}{3\times5}+\dfrac{3}{5\times7}+\dfrac{3}{7\times9}+...+\dfrac{3}{48\times50}\)
\(B=\dfrac{3}{2}\times\left(\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+\dfrac{2}{7\times9}+...+\dfrac{2}{48\times50}\right)\)
\(B=\dfrac{3}{2}\times\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{48}-\dfrac{1}{50}\right)\)
\(B=\dfrac{3}{2}\times\left(\dfrac{1}{3}-\dfrac{1}{50}\right)\)
\(B=\dfrac{3}{2}\times\dfrac{47}{150}\)
\(B=\dfrac{47}{100}\)
Chúc em học tốt!
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\(=\frac{1}{3.5}+\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{113.115}\right)\)
\(=\frac{1}{15}+\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{113}-\frac{1}{115}\right)\)
\(=\frac{1}{15}+\frac{1}{3}-\frac{1}{115}=\frac{9}{23}\)
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\(\frac{3}{5×3}+\frac{3}{5×7}+\frac{3}{7×9}+\frac{3}{9×11}+\frac{3}{11×13}\)
\(=\frac{3}{2}×\left(\frac{2}{3×5}+\frac{2}{5×7}+\frac{2}{7×9}+\frac{2}{9×11}+\frac{2}{11×13}\right)\)
\(=\frac{3}{2}×\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)
\(=\frac{3}{2}×\left(\frac{1}{3}-\frac{1}{13}\right)\)
\(=\frac{3}{2}×\left(\frac{13}{39}-\frac{3}{39}\right)\)
\(=\frac{3}{2}×\frac{10}{39}\)
\(=\frac{5}{13}\)
\(\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{7.9}+\frac{3}{9.11}+\frac{3}{11.13}\)
\(=\frac{3}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)
\(=\frac{3}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{ 1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11} +\frac{1}{11}-\frac{1}{13}\right)\)
\(=\frac{3}{2}.\left(\frac{1}{3}-\frac{1}{13}\right)\)
\(=\frac{3}{2}.\frac{10}{39}\)
\(=\frac{15}{39}\)
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\(\frac{1}{1x2} +(\frac{2}{3x5}+\frac{2}{5x7}+\frac{2}{7x9} +\frac{2}{9x11})\)
\(=\frac{1}{1x2} + (\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11})\)
\(=\frac{1}{1x2}+(\frac{1}{3}-\frac{1}{11})\)
\(=\frac{1}{1x2} +\frac{10}{33}\)
\(=\frac{1}{2} + \frac{10}{33} = \frac{33}{66}+\frac{20}{66}\)
\(=\frac{53}{66}\)
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917749738461936926399639748776398646491639394748947630373937366
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B=2/3x5 + 2/5x7 + 2/7x9 + ...+2/99x101
B= 1/3 - 1/5 + 1/5 - 1/7 + 1/7 -1/9 + ... + 1/99 - 1/101
B= 1/3 - 1/101
B=98/303
( k mk nhé ! Cách làm câu a và b của mk đều đúng 100% đấy ! Dạng này mk học từ lâu rồi ! )
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\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\)
\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{9}\right)\)
\(=\frac{1}{2}.\frac{8}{9}\)
\(=\frac{4}{9}\)
#)Giải :
\(S=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\)
\(\Rightarrow2S=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}\)
\(\Rightarrow2S=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\)
\(\Rightarrow2S=1-\frac{1}{9}=\frac{8}{9}\)
\(S=\frac{8}{9}:2=\frac{4}{9}\)
#~Will~be~Pens~#
\(\frac{3}{5\cdot7}+\frac{3}{7\cdot9}+...+\frac{3}{129\cdot131}\)
\(=3\left(\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+....+\frac{1}{129\cdot131}\right)\)
\(=\frac{3}{2}\left(\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{129\cdot131}\right)\)
\(=\frac{3}{2}\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{129}-\frac{1}{131}\right)\)
\(=\frac{3}{2}\left(\frac{1}{5}-\frac{1}{131}\right)\)
\(=\frac{3}{2}\cdot\frac{126}{655}\)
\(=\frac{378}{1310}=\frac{189}{655}\)