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\(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+...+\frac{1}{97.101}\)
\(=\frac{1}{4}\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{97}-\frac{1}{101}\right)\)
\(=\frac{1}{4}.\left(1-\frac{1}{101}\right)\)
\(=\frac{1}{4}.\frac{100}{101}\)
\(=\frac{25}{101}\)
\(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+...+\frac{1}{41.45}\)
\(=\frac{1}{4}.\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\right)\)
\(=\frac{1}{4}.\left(1-\frac{1}{45}\right)\)
\(=\frac{1}{4}.\frac{44}{45}\)
\(=\frac{11}{45}\)
Đặt \(\frac{1}{1\cdot5}+\frac{1}{5\cdot9}+\frac{1}{9\cdot13}+\frac{1}{13\cdot17}+...+\frac{1}{41\cdot45}\) là A.
Ta có:
\(A=\frac{1}{1\cdot5}+\frac{1}{5\cdot9}+\frac{1}{9\cdot13}+\frac{1}{13\cdot17}+...+\frac{1}{41\cdot45}\)
\(4A=4\left(\frac{1}{1\cdot5}+\frac{1}{5\cdot9}+\frac{1}{9\cdot13}+\frac{1}{13\cdot17}+...+\frac{1}{41\cdot45}\right)\)
\(4A=\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+...+\frac{4}{41\cdot45}\)
\(4A=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{41}-\frac{1}{45}\)
\(4A=1-\frac{1}{45}\)
\(4A=\frac{44}{45}\)
\(A=\frac{44}{45}:4\)
\(A=\frac{11}{45}\)
Vậy \(\frac{1}{1\cdot5}+\frac{1}{5\cdot9}+\frac{1}{9\cdot13}+\frac{1}{13\cdot17}+...+\frac{1}{41\cdot45}=\frac{11}{45}\)
D=1/1.5+1/5.9+...+1/41.45
4D=4/1.5+4/5.9+...+4/41.45
4D=1-1/5+1/5-1/9+...+1/41-1/45
4D=1-1/45
D=44/45:4=11/45
A= 1/5.9+1/9.13+1/13.17+1/17.21+1/21.25
4A= 4/5.9+4/9.13+4/13.17+4/17.21+4/21.25
4A= (1/5-1/9)+(1/9-1/13)+(1/13-1/17)+(1/17-1/21)+(1/21-1/25)
4A= 1/5- 1/25
4A= 4/25
A= 4/25 :4
A= 1/25
\(H=\frac{2\cdot2}{1\cdot5}+\frac{2\cdot2}{5\cdot9}+...+\frac{2\cdot2}{45.49}\)
\(H=\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+...+\frac{4}{45\cdot49}\)
\(H=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{45}-\frac{1}{49}\)
\(H=1-\frac{1}{49}\)
\(H=\frac{48}{49}\)
\(H=\frac{2.2}{1.5}+\frac{2.2}{5.9}+\frac{2.2}{9.13}+...+\frac{2.2}{45.49}\)
\(\Rightarrow H=\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{45.49}\)
\(\Rightarrow H=\frac{5-1}{1.5}+\frac{9-5}{5.9}+...+\frac{49-45}{45.49}\)
\(\Rightarrow H=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{45}-\frac{1}{49}\)
\(\Rightarrow H=1-\frac{1}{49}=\frac{48}{49}\)
a) A = 4/5.9 + 4/9.13 + 4/13.17 + ... + 4/41/45
A = 1/5 - 1/9 + 1/9 - 1/13 + 1/13 - 1/17 + ... + 1/41 - 1/45
A = 1/5 - 1/45
A = 8/45
b) B = ( 1 - 1/2 ) . ( 1 - 1/3 ) . ( 1 - 1/4 ) . ..... . ( 1 - 1/100 )
B = 1/2 . 2/3 . 3/4 . .... . 99/100
B = \(\frac{1.2.3.......99}{2.3.4......100}\)
B = 1/100
B = \(\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)+...+\left(1-\frac{1}{100}\right)\)
B = \(\frac{1}{2}.\frac{2}{3}.....\frac{99}{100}\)
B = \(\frac{1}{100}\)
\(A=\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+........+\frac{1}{100.104}\)
\(=\frac{1}{4}.\left(\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+.......+\frac{4}{100.104}\right)\)
\(=\frac{1}{4}.\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+.......+\frac{1}{100}-\frac{1}{104}\right)\)
\(=\frac{1}{4}.\left(\frac{1}{5}-\frac{1}{104}\right)\)
\(=\frac{1}{4}.\frac{99}{520}=\frac{99}{2080}\)
Thay vì nhân một số với 2008; bạn ấy nhân một số với : 2 + 0 + 0 + 8 = 10
Số đó là : 5630 / 10 = 563
\(A=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{9.11}-\frac{1}{11.13}\)
\(\Leftrightarrow A=\frac{1}{1.3}-\frac{1}{11.13}=\frac{1}{3}-\frac{1}{143}=\frac{140}{429}\)
\(A=\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}+\frac{4}{7.9.11}+\frac{4}{9.11.13}\)
\(\Rightarrow A=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{9.11}-\frac{1}{11.13}\)
\(\Rightarrow A=\frac{1}{1.3}-\frac{1}{11.13}=\frac{1}{3}-\frac{1}{143}=\frac{140}{429}\)
1/1.5 + 1/5.9 + 1/9.13 + ... + 1/97.101
= 1/4.(4/1.5 + 4/5.9 + 4/9.13 + ... + 4/97.101)
= 1/4.(1 - 1/5 + 1/5 - 1/9 + 1/9 - 1/13 + ... + 1/97 - 1/101)
= 1/4.(1 - 1/101)
= 1/4.100/101
= 25/101
\(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+........+\frac{1}{97.101}\)
\(=\frac{1}{4}\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+........+\frac{1}{97}-\frac{1}{101}\right)\)
\(=\frac{1}{4}\left(1-\frac{1}{101}\right)\)
\(=\frac{1}{4}.\frac{100}{101}=\frac{25}{101}\)