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PH
1
3 tháng 9 2017
\(F=\dfrac{49}{2.9}+\dfrac{49}{9.16}+............+\dfrac{49}{65.72}\)
\(\Leftrightarrow F=\dfrac{7^2}{2.9}+\dfrac{7^2}{9.16}+............+\dfrac{7^2}{65.72}\)
\(\Leftrightarrow F=7\left(\dfrac{7}{2.9}+\dfrac{7}{9.16}+.............+\dfrac{7}{65.72}\right)\)
\(\Leftrightarrow F=7\left(\dfrac{1}{2}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...........+\dfrac{1}{65}-\dfrac{1}{75}\right)\)
\(\Leftrightarrow F=7\left(\dfrac{1}{2}-\dfrac{1}{72}\right)\)
\(\Leftrightarrow F=7.\dfrac{35}{72}=\dfrac{245}{72}\)
\(G=\dfrac{3}{1.3}+\dfrac{3}{3.5}+...........+\dfrac{3}{47.49}\)
\(\Leftrightarrow G=\dfrac{3.2}{1.3.2}+\dfrac{3.2}{3.5.2}+........+\dfrac{3.2}{47.49}\)
\(\Leftrightarrow G=\dfrac{3}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+..........+\dfrac{2}{47.49}\right)\)
\(\Leftrightarrow G=\dfrac{3}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+........+\dfrac{1}{47}-\dfrac{1}{49}\right)\)
\(\Leftrightarrow G=\dfrac{3}{2}\left(1-\dfrac{1}{49}\right)\)
\(\Leftrightarrow G=\dfrac{3}{2}.\dfrac{48}{49}=\dfrac{72}{49}\)
24 tháng 6 2019
\(A=1.3+3.5+5.7+...+45.47+47.49\)
\(A=\left(1.49\right)+\left(2.3\right)+\left(2.5\right)+\left(2.7\right)+.....+\left(2.47\right)\)
\(A=49+2.\left(3+5+7+....+47\right)\)
Bây giờ ta phải tìm SSH của :
\(3+7+...+47\)
Vậy SSH của tổng đó là :
(47-3):2+1=23 (SSH)
=> \(A=49+2.\left(\frac{\left(47+3\right).23}{2}\right)\)
\(A=49+2.575\)
\(A=49+1150\)
\(A=1199\)
Dạng này lầm đầu gặp 
HM
11 tháng 5 2018
\(=3.\left(\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{47.49}\right)\)
\(=3.\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{47}-\dfrac{1}{49}\right)\)
\(=3.\left(\dfrac{1}{3}-\dfrac{1}{49}\right)\)
\(=3.\dfrac{46}{147}\)
\(=\dfrac{46}{49}\)
11 tháng 5 2018
\(\dfrac{3}{3.5}+\dfrac{3}{5.7}+...+\dfrac{3}{47.49}\)
=\(\dfrac{3}{2}.\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{47}-\dfrac{1}{49}\right)\)
=\(\dfrac{3}{2}.\left(\dfrac{1}{3}-\dfrac{1}{49}\right)\)
=\(\dfrac{3}{2}.\dfrac{46}{147}\)
=\(\dfrac{23}{49}\)
9 tháng 5 2018
Đặt A = \(\dfrac{3}{3.5}+\dfrac{3}{5.7}+\dfrac{3}{7.9}+...+\dfrac{3}{47.49}\)
2A = \(\dfrac{3.2}{3.5}+\dfrac{3.2}{5.7}+\dfrac{3.2}{7.9}+...+\dfrac{3.2}{47.49}\)
2A = 3\(\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{47.49}\right)\)
2A = 3 \(\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{47}-\dfrac{1}{49}\right)\)
2A = 3 \(\left(\dfrac{1}{3}-\dfrac{1}{49}\right)\)
2A = 3 . \(\dfrac{46}{147}\)
2A = \(\dfrac{46}{49}\)
=> A = \(\dfrac{46}{49}\) : 2
=> A = \(\dfrac{23}{49}\)
N
0
NV
Nguyễn Việt Lâm
Giáo viên
5 tháng 3 2023
\(2H=\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{49.51}\)
\(2H=\dfrac{3-1}{1.3}+\dfrac{5-3}{3.5}+...+\dfrac{51-49}{49.51}\)
\(2H=\dfrac{3}{1.3}-\dfrac{1}{1.3}+\dfrac{5}{3.5}-\dfrac{3}{3.5}+...+\dfrac{51}{49.51}-\dfrac{49}{49.51}\)
\(2H=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{49}-\dfrac{1}{51}\)
\(2H=1-\dfrac{1}{51}\)
\(2H=\dfrac{50}{51}\)
\(H=\dfrac{25}{51}\)
25 tháng 2 2017
\(Q=\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{47.49}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{47}-\frac{1}{49}\)
\(=\frac{1}{3}-\frac{1}{49}\)
\(=\frac{46}{147}\)
Vậy \(Q=\frac{46}{147}\)
25 tháng 2 2017
Ta có : \(\frac{2}{3}Q=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{47.49}\)
\(\Rightarrow\frac{2}{3}Q=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{47}-\frac{1}{49}\)
\(\Rightarrow\frac{2}{3}Q=\frac{1}{3}-\frac{1}{49}=\frac{49}{147}-\frac{3}{147}=\frac{46}{147}\)
\(\Rightarrow Q=\frac{46}{147}\div\frac{2}{3}=\frac{138}{294}=\frac{23}{49}\)
Vậy ...
1/1-1/3+1/3-1/5+1/5-1/7+...... +1/47-1/49
3/1.3+3/3.5+3/5.7+......+3/47.49
=1/1-1/3+1/3-1/5+1/5-1/7+........+1/47-1/49
=1/1-1/49
=49/49-1/49
=48/49