\(\frac{5}{4x7}\)+\(\frac{5}{7x10}\)+\(\frac{5}{10x13}\)+....+\(\frac{5}{301x304}\)
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Ta có:
A = \(\frac{5}{4.7}+\frac{5}{7.10}+\frac{5}{10.13}+...+\frac{5}{301.304}\)
A = 5. (\(\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{301.304}\))
3A = 3.5. (\(\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{301.304}\))
3A = 5. (\(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{301.304}\))
3A = 5. ( \(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-...-\frac{1}{301}+\frac{1}{301}-\frac{1}{304}\))
3A = 5. ( \(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-...-\frac{1}{301}+\frac{1}{301}-\frac{1}{304}\))
3A = 5. ( \(\frac{1}{4}-\frac{1}{304}\))
3A = \(\frac{5.75}{304}\)
3A = \(\frac{375}{304}\)
A= \(\frac{125}{304}\) . Vậy: A = \(\frac{125}{304}\)
Chúc bạn học tốt! Tick cho mình nhé!
E= 7/4x7 + 7/7x10 =7/10x13+...+ 7/301x304
\(=\frac{7}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{301}-\frac{1}{304}\right)\)
\(=\frac{7}{3}\left(\frac{1}{4}-\frac{1}{304}\right)\)
\(=\frac{7}{3}\cdot\frac{75}{304}\)
\(=\frac{175}{304}\)
E = \(\frac{7}{4.7}+\frac{7}{7.10}+\frac{7}{10.13}+...+\frac{7}{301.304}\)
=\(\frac{7}{3}.\frac{7-4}{4.7}+\frac{7}{3}.\frac{10-7}{7.10}+\frac{7}{3}.\frac{13-10}{10.13}+...+\frac{7}{3}.\frac{304-301}{301.304}\)
= \(\frac{7}{3}.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{301}-\frac{1}{304}\right)\)=\(\frac{7}{3}.\left(\frac{1}{4}-\frac{1}{304}\right)=\frac{7}{3}.\frac{75}{304}=\frac{175}{304}\)
=3/1-3/4+3/4-3/7+3/7-3/10+3/10-3/13
=1/1-1/4+1/4-1/7+1/7-1/10+1/10-1/13
=1-1/13
=12/13
\(D=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\)
\(D=1-\frac{1}{100}\)
\(D=\frac{99}{100}\)
1/1*4 + 1/4*7 + 1/7*10 + ... + 1/97*100
= 1/3(3/1*4 + 3/4*7 + 3/7*10 + ... + 3/97*100)
= 1/3(1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + .... + 1/97 - 1/100)
= 1/3(1 - 1/100)
= 1/3*99/100
= 33/100
Q=\(\frac{1}{2}-\frac{1}{9}+\frac{1}{9}-\frac{1}{7}+\frac{1}{7}-\frac{1}{19}+...+\frac{1}{252}-\frac{1}{509}\)
=\(\frac{1}{2}-\left(\frac{1}{9}+\frac{1}{9}\right)-\left(\frac{1}{7}+\frac{1}{7}\right)-...-\left(\frac{1}{252}+\frac{1}{252}\right)-\frac{1}{509}\)
=\(\frac{1}{2}-0+0+0+...+0-\frac{1}{509}\)
=\(\frac{1}{2}-\frac{1}{509}\)
=\(\frac{507}{1018}\)
MẤY CÂU KHÁC THÌ TƯƠNG TỰ, CHÚC BẠN MAY MẮN!!!:))
\(D=\frac{2}{1\cdot4}+\frac{2}{4\cdot7}+\frac{2}{7\cdot10}+...+\frac{2}{97\cdot100}\)
\(D=\frac{2}{3}\left(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{97\cdot100}\right)\)
\(D=\frac{2}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(D=\frac{2}{3}\left(1-\frac{1}{100}\right)\)
\(D=\frac{2}{3}\cdot\frac{99}{100}=\frac{33}{50}\)
\(=\frac{5}{3}.\left(\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{301.304}\right)\)
\(=\frac{5}{3}.\left(\frac{1}{4}-\frac{1}{304}\right)\)
\(=\frac{5}{3}.\frac{75}{304}\)
\(=\frac{125}{304}\)
\(\frac{5}{4×7}+\frac{5}{7×10}+\frac{5}{10×13}+...+\frac{5}{301×304}\)
\(=\frac{5}{4}-\frac{5}{7}+\frac{5}{7}-\frac{5}{10}+\frac{5}{10}-\frac{5}{13}+...+\frac{5}{301}-\frac{5}{304}\)
\(=\frac{5}{4}-\frac{5}{304}\)
\(=\frac{380}{304}-\frac{5}{304}\)
\(=\frac{375}{304}\)
Cbht