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\(\frac{1}{2}+\frac{1}{14}+\frac{1}{35}+\frac{1}{65}+\frac{1}{104}+\frac{1}{152}\)
\(=\frac{1}{1.2}+\frac{1}{2.7}+\frac{1}{7.5}+\frac{1}{5.13}+\frac{1}{13.8}+\frac{1}{8.19}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{5}+\frac{1}{5}-\frac{1}{13}+\frac{1}{13}-\frac{1}{8}+\frac{1}{8}-\frac{1}{19}\)
\(=1-\frac{1}{19}\)
\(=\frac{18}{19}\)
~Học tốt~
\(\frac{1}{2}+\frac{1}{14}+\frac{1}{35}+\frac{1}{65}+\frac{1}{104}+\frac{1}{152}\)
\(=\frac{1}{2}+\frac{1}{2.7}+\frac{1}{7.5}+\frac{1}{5.13}+\frac{1}{13.8}+\frac{1}{8.19}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{5}+\frac{1}{5}-\frac{1}{13}+\frac{1}{13}-\frac{1}{8}+\frac{1}{8}-\frac{1}{19}\)
\(=1-\frac{1}{19}\)
\(=\frac{18}{19}\)
\(\frac{1}{2}+\frac{1}{14}+\frac{1}{35}+\frac{1}{65}+\frac{1}{104}+\frac{1}{152}=\frac{12}{19}\)
Ai đồng ý với ý kiến của mk thì , mk sẽ tích lại gấp đôi!!!!!!
Bạn vô câu hỏi tương tự của bạn trần vũ quang anh là: olm.vn/hoi_dap/question/76402.html
Bài đó bạn nguyễn thị ngọc anh làm đúng 100% luôn đó
k cho mik nhé
\(\frac{1}{2}+\frac{1}{14}+\frac{1}{35}+\frac{1}{65}+\frac{1}{104}+\frac{1}{152}\)
\(=2.\left(\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+\frac{1}{130}+\frac{1}{208}+\frac{1}{304}\right)\)
\(=2.\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+\frac{1}{13.16}+\frac{1}{16.19}\right)\)
\(=2.\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{16.19}\right)\)
\(=\frac{2}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{16}-\frac{1}{19}\right)\)
\(=\frac{2}{3}.\left(1-\frac{1}{19}\right)\)
\(=\frac{2}{3}.\frac{18}{19}=\frac{12}{19}\)
=1/1.2+1/2.7+1/7.5+1/5.13+1/13.8+1/8.19
=2/1.4+2/4.7+2/7.10+2/10.13+2/13.16+2/16.19(ca tu va mau cung nhan them 2 gia tri khong doi)
=2/3.(3/1.4+3/4.7+3/7.10+3/10.13+3/13.16+3/16.19)
=2/3.(1/1-1/4+1/4-1/7+1/7-1/10+1/101/13+1/13-1/16+1/16-1/19)
=2/3.18/19=12/19
Ta có:
\(\frac{1}{2} + \frac{1}{14} + \frac{1}{35} + \frac{1}{65} + \frac{1}{104} + \frac{1}{152}\)
Quy đồng mẫu số chung là:
\(\text{MSC} = 69160\)
Khi đó:
\(\frac{1}{2}=\frac{34580}{69160},\frac{1}{14}=\frac{4940}{69160},\frac{1}{35}=\frac{1976}{69160},\frac{1}{65}=\frac{1064}{69160},\frac{1}{104}=\frac{665}{69160},\frac{1}{152}=\frac{455}{69160}.\)
Vậy:
\(\frac{43680}{69160} = \frac{12}{19} .\)
\(\boxed{\star19/08/2004}\)
Đặt \(A=\dfrac12+\dfrac{1}{14}+\dfrac{1}{35}+\dfrac{1}{65}+\dfrac{1}{104}+\dfrac{1}{152}\)
\(\Rightarrow\dfrac12A=\dfrac12\left(\dfrac12+\dfrac{1}{14}+\dfrac{1}{35}+\dfrac{1}{65}+\dfrac{1}{104}+\dfrac{1}{152}\right)\)
\(\Rightarrow\dfrac12A=\dfrac14+\dfrac{1}{28}+\dfrac{1}{70}+\dfrac{1}{130}+\dfrac{1}{208}+\dfrac{1}{304}\)
\(\Rightarrow\dfrac12A=\dfrac{1}{1\cdot4}+\dfrac{1}{4\cdot7}+\dfrac{1}{7\cdot10}+\dfrac{1}{10\cdot13}+\dfrac{1}{13\cdot16}+\dfrac{1}{16\cdot19}\)
\(\Rightarrow3\cdot\dfrac12A=\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+\dfrac{3}{10\cdot13}+\dfrac{3}{13\cdot16}+\dfrac{3}{16\cdot19}\)
\(\Rightarrow\dfrac32A=\left(1-\dfrac14\right)+\left(\dfrac14-\dfrac17\right)+\left(\dfrac17-\dfrac{1}{10}\right)+\left(\dfrac{1}{10}-\dfrac{1}{13}\right)+\left(\dfrac{1}{13}-\dfrac{1}{16}\right)+\left(\dfrac{1}{16}-\dfrac{1}{19}\right)\)
\(\Rightarrow\dfrac32A=1-\dfrac{1}{19}\)
\(\Rightarrow\dfrac32A=\dfrac{18}{19}\)
\(\Rightarrow A=\dfrac{18}{19}:\dfrac32\)
\(\Rightarrow A=\dfrac{18}{19}\cdot\dfrac23\)
\(\Rightarrow A=\dfrac{12}{19}\)