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\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
\(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}+\frac{2}{99.101}\)
\(=1-\frac{1}{3}+\frac{!}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}=\frac{100}{101}\)
\(C=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+....+\frac{1}{1024}+\frac{1}{2048}\)
\(\Rightarrow\)\(2C=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+....+\frac{1}{512}+\frac{1}{1024}\)
\(\Rightarrow\)\(2C-C=\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1024}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2048}\right)\)
\(\Leftrightarrow\)\(C=1-\frac{1}{2048}=\frac{2047}{2048}\)
Ta có :\(B=\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.\frac{4^2}{4.5}.....\frac{98^2}{98.99}=\frac{\left(1.2.3.4...98\right).\left(1.2.3.4...98\right)}{\left(1.2.3.4...98\right).\left(2.3.4.5...99\right)}=\frac{1}{99}\)
Lại có A = \(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}=1-\frac{1}{99}=\frac{98}{99}\)
Lại có \(A:B=\frac{98}{99}:\frac{1}{99}=98\)
=> A = 98B
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
a) \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{2017\cdot2018}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(=1-\frac{1}{2018}\)
\(=\frac{2017}{2018}\)
b) \(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{97\cdot99}\)( sửa 91.99 thành 97.99 mới đúng nha )
\(=\frac{1}{2}\left(\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+...+\frac{2}{97}-\frac{2}{99}\right)\)
\(=\frac{1}{2}\left(\frac{2}{3}-\frac{2}{99}\right)\)
\(=\frac{1}{2}.\frac{64}{99}\)
\(=\frac{32}{99}\)
a) 1/1.2 + 1/2.3 + 1/3.4 +...+1/2017.2018
= 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ....+1/2017 - 1/2018
= 1 - 1/2018
= 2017/2018
a) \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
b) \(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
\(=2.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=2.\left(1-\frac{1}{99}\right)\)
\(=2.\frac{98}{99}\)
\(=\frac{196}{99}=1\frac{97}{99}\)
Đây là toán lớp 1 hả??????????? Tớ nghĩ là toán lớ 6 đấy!!!!!!
a) \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{99\cdot100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
b) \(\frac{2}{3\cdot5}+\frac{3}{5\cdot7}+...+\frac{2}{49\cdot51}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\)
\(=\frac{1}{3}-\frac{1}{51}\)
\(=\frac{16}{51}\)
a) 1/1.2+1/2.3+1/3.4+...+1/99.100
= 1/1 - 1/2 + 1/2 - 1/3 + 1/3 -1/4 + ... + 1/99 - 1/100
= 1/1 - 1/100
= 99/100
b) 2/3.5+2/5.7+...+2/49.51
= 2 . ( 1/3.5 + 1/5.7 + ... + 1/49.51 )
= 2 . ( 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/49 - 1/50 )
= 2 . ( 1/3 - 1/50 )
= 2 . 47/150
= 47/75
Bài 1:
a) A=1/1.2+1/2.3+1/3.4+...+1/98.99+1/99.100
A=1-1/2+1/2-1/3+1/3-1/4+...+1/98-1/99+1/99-1/100
A=1-1/100
A=99/100
b) B=1/3.5+1/5.7+1/7.9+...+1/145.147
Ta nhân biểu thức trên với 2 ta được
B=2/3.5+2/5.7+2/7.9+...+2/145.147
B=2.(1/3.5+1/5.7+1/7.9+...+1/145.147)
B=2.(1/3-1/5+1/5-1/7+1/7-1/9+...+1/145-1/147)
B=2.(1/3-1/147)
B=2.16/49
B=32/49
Bài 2:
Ta có: 4/11<x/20<5/11
=> 88/220<11x/220<100/220
=> 80<11x<100
=> 11x thuộc {88;99}
=> x thuộc {8;9}
bạn thuysvaan ơi làm lại câu A đc ko