tính hợp lí:
\(\frac{2}{2x3x4}\)+ \(\frac{2}{3x4x5}\)+ ............+\(\frac{2}{37x38x39}\)+ \(\frac{2}{38x39x40}\)
ai đúng mk tích cho 5 tick
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\(A=\frac{1}{6.10}+\frac{1}{10.14}+\frac{1}{14.18}+\frac{1}{18.22}+\frac{1}{22.26}+\frac{1}{26.30}\)
\(=\frac{1}{4}.\left(\frac{1}{6}-\frac{1}{10}+\frac{1}{10}-\frac{1}{14}+\frac{1}{14}-\frac{1}{18}+\frac{1}{18}-\frac{1}{22}+\frac{1}{22}-\frac{1}{26}+\frac{1}{26}-\frac{1}{30}\right)\)
\(=\frac{1}{4}.\left(\frac{1}{6}-\frac{1}{30}\right)=\frac{1}{4}.\frac{2}{15}=\frac{1}{30}\)
\(B=\frac{5}{2.3}+\frac{5}{3.4}+\frac{5}{4.5}+...+\frac{5}{8.9}\)\(=5.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{8.9}\right)\) \(=5.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{8}-\frac{1}{9}\right)\)
\(=5.\left(\frac{1}{2}-\frac{1}{9}\right)=5.\frac{7}{18}=\frac{35}{18}\)
\(C=\left(\frac{7^2}{2.9}+\frac{7^2}{9.16}+....+\frac{7^2}{65.72}\right):\left(\frac{1}{3}-\frac{7}{36}\right)\)
\(=7.\left(\frac{7}{2.9}+\frac{7}{9.16}+...+\frac{7}{65.72}\right):\frac{5}{36}\) \(=7.\left(\frac{1}{2}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+...+\frac{1}{65}-\frac{1}{72}\right):\frac{5}{36}\)'
\(=7.\left(\frac{1}{2}-\frac{1}{72}\right):\frac{5}{36}=7.\frac{35}{72}:\frac{5}{36}=\frac{49}{2}\)
\(D=\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{37.38.39}+\frac{2}{38.39.40}\)
\(=2.\left(\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{37.38.39}+\frac{1}{38.39.40}\right)\)
\(=2.\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{37.38}-\frac{1}{38.39}+\frac{1}{38.39}-\frac{1}{39.40}\right)\)
\(=\frac{1}{2.3}-\frac{1}{39.40}=\frac{259}{1560}\)
\(E=\frac{202202}{1212}+\frac{202202}{2020}+\frac{202202}{3030}+\frac{202202}{4242}+\frac{202202}{5656}\)
\(=202202.\left(\frac{1}{3.4.101}+\frac{1}{4.5.101}+\frac{1}{5.6.101}+\frac{1}{6.7.101}+\frac{1}{7.8.101}\right)\)
\(=2002.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\right)\)
\(=2002.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\right)\)
\(=2002.\left(\frac{1}{3}-\frac{1}{8}\right)=2002.\frac{5}{24}=\frac{5005}{12}\)
\(A=\frac{1}{1\times2\times3}+\frac{1}{2\times3\times4}+\frac{1}{3\times4\times5}+...+\frac{1}{36\times37\times38}+\frac{1}{37\times38\times39}\)
\(2A=\frac{2}{1\times2\times3}+\frac{2}{2\times3\times4}+\frac{2}{3\times4\times5}+...+\frac{2}{36\times37\times38}+\frac{2}{37\times38\times39}\)
\(2A=\frac{1}{1\times2}-\frac{1}{2\times3}+\frac{1}{2\times3}-\frac{1}{3\times4}+...+\frac{1}{37\times38}-\frac{1}{38\times39}\)
\(2A=\frac{1}{1\times2}-\frac{1}{38\times39}\)
\(2A=\frac{741}{1482}-\frac{1}{1482}\)
\(2A=\frac{370}{741}\)
\(A=\frac{370}{741}:2=\frac{185}{741}\)
Sửa đề chút:
\(\frac{1}{1x2x3}+\frac{1}{2x3x4}+...+\frac{1}{98x99x100}\)
\(=\frac{1}{2}.\left(\frac{2}{1x2x3}+\frac{2}{2x3x4}+...+\frac{2}{98x99x100}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1x2}-\frac{1}{2x3}+\frac{1}{2x3}-\frac{1}{3x4}+...+\frac{1}{98x99}-\frac{1}{99x100}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{99.100}\right)\)
\(=\frac{1}{4}-\frac{1}{99.200}< 1\)
đpcm
\(2C=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{39-37}{37.38.39}\)
\(2C=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{37.38}-\frac{1}{38.39}\)
\(2C=\frac{1}{1.2}-\frac{1}{38.39}\)
\(C=\frac{617}{1482}\)
\(3D=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\)
\(3D-D=1-\frac{1}{3^8}\)
\(D=\frac{1}{2}-\frac{1}{2.3^8}\)
Ta có:\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{37.38}-\frac{1}{38.39}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{38.39}\right)\)
b,\(D=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^8}\)
\(\Rightarrow3D=1+\frac{1}{3}+\frac{1}{3^2}+.....+\frac{1}{3^7}\)
\(\Rightarrow2D=1-\frac{1}{3^8}\)
\(\Rightarrow D=\frac{3^8-1}{3^8}:2\)
Bài làm
\(2\frac{5}{12}-99\frac{213}{9115}:\frac{12}{5}+98\frac{213}{9115}:\frac{12}{5}\)
\(=2\frac{5}{12}-\left(99\frac{213}{9115}+98\frac{214}{9115}\right):\frac{12}{5}\)
\(=2\frac{5}{12}-197:\frac{12}{5}\)
\(=2\frac{5}{12}-197.\frac{5}{12}\)
\(=2\frac{5}{12}-197\frac{5}{12}\)
\(=\left(2-197\right)\frac{5}{12}\)
\(=-195\frac{5}{12}\)
\(=-2335\)
~ Chắc đúng ~
# Chúc bạn học tốt #
Bài làm
Mik sai mất rồi. Mik sửa nè:
\(=-195\frac{5}{12}\)
\(=-\frac{2335}{12}\)
~ Thế mới đúng nha ~
# Chúc bạn học tốt #
Ta có: \(\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{37.38.39}\)
\(\Rightarrow\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+.....+\frac{1}{37}-\frac{1}{38}+\frac{1}{38}-\frac{1}{39}\)
\(=\frac{1}{2.3}-\frac{1}{38.39}=\frac{41}{247}\)