\(\text{5/4x7+5/7x10+5/10x13+⋯+5/58x61}\)
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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!
\(=\frac{5}{3}.\left(\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{301.304}\right)\)
\(=\frac{5}{3}.\left(\frac{1}{4}-\frac{1}{304}\right)\)
\(=\frac{5}{3}.\frac{75}{304}\)
\(=\frac{125}{304}\)
\(\frac{5}{4×7}+\frac{5}{7×10}+\frac{5}{10×13}+...+\frac{5}{301×304}\)
\(=\frac{5}{4}-\frac{5}{7}+\frac{5}{7}-\frac{5}{10}+\frac{5}{10}-\frac{5}{13}+...+\frac{5}{301}-\frac{5}{304}\)
\(=\frac{5}{4}-\frac{5}{304}\)
\(=\frac{380}{304}-\frac{5}{304}\)
\(=\frac{375}{304}\)
Cbht
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 = \(\frac{7}{4.7}+\frac{7}{7.10}+\frac{7}{10.13}+...+\frac{7}{301.304}\)
=\(\frac{7}{3}.\frac{7-4}{4.7}+\frac{7}{3}.\frac{10-7}{7.10}+\frac{7}{3}.\frac{13-10}{10.13}+...+\frac{7}{3}.\frac{304-301}{301.304}\)
= \(\frac{7}{3}.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{301}-\frac{1}{304}\right)\)=\(\frac{7}{3}.\left(\frac{1}{4}-\frac{1}{304}\right)=\frac{7}{3}.\frac{75}{304}=\frac{175}{304}\)
\(C=\dfrac{1}{4.7}+\dfrac{1}{7.10}+\dfrac{1}{10.13}+...+\dfrac{1}{2020+2023}\)
\(=\dfrac{1}{3}\left(\dfrac{3}{4.7}+\dfrac{3}{7.10}+\dfrac{3}{10.13}+...+\dfrac{3}{2020.2023}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+...+\dfrac{1}{2020}-\dfrac{1}{2023}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{4}-\dfrac{1}{2023}\right)\)
\(=\dfrac{1}{3}.\dfrac{2019}{8092}\)
\(=\dfrac{673}{8092}\)
=1/1-1/4+1/4-1/7+1/7-1/10+1/10-1/13+1/13-1/16
=1-1/16=15/16
1/1.4+1/4.7+1/7.10+1/10.13+1/13.16
=1/3.(3/1.4+3/4.7+3/7.10+3/10.13+3/13.16)
=1/3.(1/1-1/4+1/4-1/7+1/7-1/10+1/10-1/13+1/13-1/16)
=1/3.(1/1-1/16)
=1/3.(16/16-1/16)=1/3.15/16=5/16
3/(1×4)+3/(4×7)+3/(7×10)+3/(10×13)+3/(13×16)
=1-1/4+1/4-1/7+1/7-1/10+1/10-1/13+1/13-1/16
=1-1/16
=15/16
31 x 434 x 737 x 10310 x 13 = 1.3289876e+12
mik phải dùng máy tính chứ có sịp nhân mới trả lời đc
nhỉ ?????
\(\dfrac{5}{4\text{x}7}+\dfrac{5}{7\text{x}10}+...+\dfrac{5}{58\text{x}61}\)
\(=\dfrac{5}{3}\text{x}\left(\dfrac{3}{4\text{x}7}+\dfrac{3}{7\text{x}10}+...+\dfrac{3}{58\text{x}61}\right)\)
\(=\dfrac{5}{3}\text{x}\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{58}-\dfrac{1}{61}\right)\)
\(=\dfrac{5}{3}\text{x}\left(\dfrac{1}{4}-\dfrac{1}{61}\right)=\dfrac{5}{3}\text{x}\dfrac{57}{244}=\dfrac{95}{244}\)
\(\dfrac{5}{4\times7}+\dfrac{5}{7\times10}+\dfrac{5}{10\times13}+...+\dfrac{5}{58\times61}\\ =\dfrac{5}{3}\times\left(\dfrac{3}{4\times7}+\dfrac{3}{7\times10}+\dfrac{3}{10\times13}+...+\dfrac{3}{58\times61}\right)\\ =\dfrac{5}{3}\times\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+...+\dfrac{1}{58}-\dfrac{1}{61}\right)\\ =\dfrac{5}{3}\times\left(\dfrac{1}{4}-\dfrac{1}{61}\right)\\ =\dfrac{5}{3}\times\dfrac{57}{244}=\dfrac{95}{244}\)