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A=3/1.3+3/3.5+3/5.7+............+3/49.51
A=3/1-3/3=3/3-3/5+3/5-3/7+...............+3/49-3/51
A=1-1/3+1/3-1/5+1/5-1/7+.....................+1/39-1/51
A=1-1/51
A=50/51
A\(=3\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...\frac{1}{49.51}\right) \)
\(=\frac{3}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...\frac{2}{49.51}\right)\)
\(=\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)
=\(\frac{3}{2}\left(1-\frac{1}{51}\right)\)
\(=\frac{3}{2}.\frac{50}{51}\)
\(=\frac{25}{17}\)
Ta có :
\(A=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{49.51}\)
\(A=\frac{3}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\right)\)
\(A=\frac{3}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(A=\frac{3}{2}\left(1-\frac{1}{51}\right)\)
\(A=\frac{3}{2}.\frac{50}{51}\)
\(A=\frac{25}{17}\)
Vậy \(A=\frac{25}{17}\)
Chúc bạn học tốt ~
\(A=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{49.51}\)
\(A=\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(A=\frac{3}{2}\left(1-\frac{1}{51}\right)\)
\(A=\frac{3}{2}.\frac{50}{51}\)
\(A=\frac{25}{17}\)
\(B=\frac{21}{4}\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{3333}{4242}\right)\)
\(B=\frac{21}{4}\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)
\(B=\frac{21}{4}\left(\frac{33}{3.4}+\frac{33}{4.5}+\frac{33}{5.6}+\frac{33}{6.7}\right)\)
\(B=\frac{21}{4}.33.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(B=\frac{21}{4}.33.\left(\frac{1}{3}-\frac{1}{7}\right)\)
\(B=\frac{21}{4}.33.\frac{4}{21}\)
\(B=\left(\frac{21}{4}.\frac{4}{21}\right).33\)
\(B=33\)
\(C=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)
\(C=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(C=\frac{1}{2}\left(1-\frac{1}{99}\right)\)
\(C=\frac{1}{2}.\frac{98}{99}\)
\(C=\frac{49}{99}\)
\(C=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{99.101}\)
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-...-\frac{1}{99}+\frac{1}{99}-\frac{1}{101}\)
\(=\frac{1}{1}-\frac{1}{101}\)
\(=\frac{100}{101}\)
\(C=\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+......+\frac{1}{99}-\frac{1}{101}\right)\)
\(C=\frac{3}{2}\left(1-\frac{1}{101}\right)\)
\(C=\frac{3}{2}.\frac{100}{101}=\frac{150}{101}\)
=3(1/1.3+1/3.5+1/5.7+1/7.9)
=3/2(1-1/3+1/3-1/5+1/5-1/7+1/7-1/9) vi khoang cach tu 1-3;3-5;5-7;7-9 la 2 nen ta nhan tat ca voi 1/2 ma 3.1/2=3/2
=3/2.(1-1/9) rut gon -1/3+1/3;-1/5+1/5;-1/7+1/7=0
=3/2.8/9=4/3
ta có :3/(1.3)+3/(3.5)+3/(5.7)+3/(7.9)
ta đặt 3 làm chung rồi tự làm đc
\(A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{51}\)
\(A=1-\frac{1}{51}\)
\(A=\frac{50}{51}\)
\(A=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{49.51}\)
\(2A=3\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\right)\)
\(2A=3\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(2A=3\left(1-\frac{1}{51}\right)\)
\(2A=3.\frac{50}{51}\)
\(2A=\frac{50}{17}\Rightarrow A=\frac{25}{17}\)'
A=3/2(2/1.3+2/3.5+2/5.7+....+2/53.55)
=3/2(1-1/3+1/3-1/5+1/5-1/7+..../1/53-1/55)
=3/2(1-1/55)
=3/2.54/55
=81/55
\(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+....+\frac{3}{49.51}\)
\(=\frac{3}{2}\left(1-\frac{1}{3}+\frac{!}{3}-\frac{1}{5}+....+\frac{1}{49}-\frac{1}{51}\right)\)
\(=\frac{3}{2}\left(1-\frac{1}{51}\right)\)
\(=\frac{3}{2}.\frac{50}{51}=\frac{25}{17}\)
Đặt \(\)A = dãy trên
Ta có \(\frac{2}{3}A=\frac{2}{3}.\left(\frac{3}{1.3}+\frac{3}{3.5}+...+\frac{3}{49.51}\right)\)
\(=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{49.51}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{51}\)
\(=1-\frac{1}{51}\)
\(=\frac{50}{51}\)
\(\Rightarrow A=\frac{50}{51}\div\frac{2}{3}=\frac{25}{17}\)
Vậy kq của dãy là\(\frac{25}{17}\)
Ta có :
\(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{99.101}\)
\(=\)\(\frac{3}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)\)
\(=\)\(\frac{3}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\)\(\frac{3}{2}\left(1-\frac{1}{101}\right)\)
\(=\)\(\frac{3}{2}.\frac{100}{101}< \frac{3}{2}.\frac{100}{100}=1,5\)
\(\Rightarrow\)\(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{99.101}< 1,5\) ( đpcm )
Vậy \(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{99.101}< 1,5\)
Chúc bạn học tốt ~
tham khảo ở đây nhé
https://olm.vn/hoi-dap/question/185679.html
^^
\(D=\frac{3}{1.3}+\frac{3}{3.5}+...+\frac{3}{199.201}\)
\(D=\frac{3}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{199.201}\right)\)
\(D=\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{199}-\frac{1}{201}\right)\)
\(D=\frac{3}{2}\left(1-\frac{1}{201}\right)\)
\(D=\frac{3}{2}.\frac{200}{201}\)
\(D=\frac{100}{67}\)
#)Giải :
\(D=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{199.201}\)
\(D=\frac{3}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{199.201}\right)\)
\(D=\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{199}-\frac{1}{201}\right)\)
\(D=\frac{3}{2}\left(1-\frac{1}{201}\right)\)
\(D=\frac{3}{2}\times\frac{200}{201}\)
\(D=\frac{100}{67}\)