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tớ làm câu b thôi, câu a nhân 1/2 lên là đc
\(A=\frac{1}{2}.\left[\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{\left(2n-1\right).\left(2n+1\right)}\right)\right]\)
\(A=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2.n-1}-\frac{1}{2n+1}\right)\)
\(A=\frac{1}{2}.\left(1-\frac{1}{2n+1}\right)=\frac{1}{2}-\frac{1}{2.\left(2n+1\right)}< \frac{1}{2}\)
p/s: lưu ý không có dấu "=" đâu nhé vì \(\frac{1}{2.\left(2n+1\right)}>0\left(n\text{ thuộc }N\right)\)
Đặt tên bthuc là A
\(A=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{19.21}\)
\(2A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{19.21}\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{21}\)
\(2A=1-\frac{1}{21}=\frac{20}{21}\)
=>\(A=\frac{20}{21}:2=\frac{10}{21}\)
\(=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{17.19}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{17}-\frac{1}{19}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{19}\right)=\frac{1}{2}.\left(\frac{18}{19}\right)\)
\(=\frac{9}{19}\)
gọi biểu thức là A
ta có :
A = \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}...\frac{1}{19.21}\)
=> 2A = \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}...\frac{2}{19.21}\)
2A = \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...-\frac{1}{21}\)
2A = 1 - \(\frac{1}{21}\)
2A = \(\frac{20}{21}\)
A = \(\frac{20}{21}:2=\frac{10}{21}\)
\(\text{Đ}\text{ặt}:A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+..+\frac{1}{99.101}\)
\(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(2A=1-\frac{1}{101}\)
\(A=\frac{100}{101}:2=\frac{50}{101}\)
\(\Rightarrow\frac{1}{3}x.x=\frac{50}{101}\)
\(x.\left(\frac{1}{3}.1\right)=\frac{50}{101}\)
\(x.\frac{1}{3}=\frac{50}{101}\)
$x=\frac{50}{101}:\frac{1}{3}=\frac{150}{101}$
\(.\frac{1}{3}x.x=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(\frac{1}{3}xx=\frac{1}{2}.\left(1-\frac{1}{101}\right)\)
\(\frac{1}{3}xx=\frac{1}{2}.\left(\frac{100}{101}\right)\)
\(\frac{1}{3}xx=\frac{50}{101}\)
\(x.x=\frac{150}{101}\)
còn lại tự tính
A\(A=\frac{1}{1.3}+..+\frac{1}{x\left(x+1\right)}\)
\(2A=\frac{1}{1}-\frac{1}{\left(x+1\right)}\)
\(A=\frac{x}{2.\left(x+1\right)}=\frac{8}{17}=\frac{16}{2.17}\)
X=16
\(A=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)
\(A=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)-\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\right)\)
\(A=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}\right)-\frac{1}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}\right)\)
\(A=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}\right)\)
\(A=\frac{1}{2}\left(1-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(A=\frac{4}{9}-\frac{1}{5}=\frac{11}{45}\)
\(S=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)
\(S=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)-\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\right)\)
\(S=\frac{1}{2}\left(1-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{10}\right)\)
\(S=\frac{1}{2}\left(1-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(S=\frac{1}{2}.\frac{8}{9}-\frac{1}{2}.\frac{2}{5}\)
\(S=\frac{4}{9}-\frac{1}{5}\)
\(S=\frac{11}{45}\)
Ta có:\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{19.21}=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{19.21}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}\right)=\frac{1}{2}\left(1-\frac{1}{21}\right)=\frac{1}{2}.\frac{20}{21}=\frac{10}{21}\)
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}\)\(+...+\frac{1}{19.21}\)
=\(\frac{2}{2}\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{19.21}\right)\)
=\(\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{19.21}\right)\)
=\(\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{19}-\frac{1}{21}\right)\)
=\(\frac{1}{2}\left(\frac{1}{1}-\frac{1}{21}\right)\)
=\(\frac{1}{2}.\frac{20}{21}\)
=\(\frac{20}{42}=\frac{10}{21}\)
=\(\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{2.4}+...+\frac{2}{8.10}\right)\)
= \(\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+....+\frac{1}{8}-\frac{1}{10}\right)\)
= \(\frac{1}{2}.\left(1+\frac{1}{2}-\frac{1}{9}-\frac{1}{10}\right)\)
=\(\frac{29}{45}\)
\(M=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2017.2019}\)
\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2017}-\frac{1}{2019}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{2019}\right)\)
\(=\frac{1}{2}.\frac{2018}{2019}\)
\(=\frac{2018}{4038}\)
\(\Rightarrow\frac{2018}{4038}< \frac{1}{2}\)( lấy máy tính )
\(M=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+.....+\frac{1}{2017.2019}\)
\(\Rightarrow M=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-......-\frac{1}{2017}+\frac{1}{2017}-\frac{1}{2019}\)
\(\Rightarrow M=1-\frac{1}{2019}\)
\(\Rightarrow M=\frac{2019}{2019}-\frac{1}{2019}\)
\(\Rightarrow M=\frac{2018}{2019}\)
Có \(\frac{2018}{2019}=\frac{2018.2}{2019.2}=\frac{4036}{4038}\)
\(\frac{1}{2}=\frac{1.2019}{2.2019}=\frac{2019}{4038}\)
Mà \(\frac{4036}{4038}< \frac{2019}{4038}\Rightarrow M< \frac{1}{2}\)
Vậy M < \(\frac{1}{2}\)