Bai nay mk vua giai xong nen ko muon giai nua

mk chi neu cach giai thoi

dau tien tinh 1/2P ra quy luat o mau la 2.6 ; 6.10 ; 10.14 ; ...; 198.202 roi nhan voi 4 la ra

Ket qua la 25/101

kết quả là \(\frac{25}{101}\)

\(P=\frac{1}{2.3}+\frac{1}{6.5}+............+\frac{1}{198.101}\)

\(P=\frac{1}{6}+\frac{1}{30}+.........+\frac{1}{19998}\)

\(\frac{1}{2}P=\frac{1}{12}+\frac{1}{60}+...........+\frac{1}{39996}\)

\(\frac{1}{2}P=\frac{1}{2.6}+\frac{1}{6.10}+.........+\frac{1}{198.202}\)

\(\frac{1}{2}P=\frac{1}{4}.\left(\frac{4}{2.6}+\frac{4}{6.10}+..........+\frac{4}{198.202}\right)\)

\(\frac{1}{2}P=\frac{1}{4}.\left(\frac{1}{2}-\frac{1}{6}+\frac{1}{6}-\frac{1}{10}+..........+\frac{1}{198}-\frac{1}{202}\right)\)

\(\frac{1}{2}P=\frac{1}{4}.\left(\frac{1}{2}-\frac{1}{202}\right)\)

\(\frac{1}{2}P=\frac{1}{4}.\frac{50}{101}\)

\(\frac{1}{2}P=\frac{25}{202}\)

\(P=\frac{25}{202}:\frac{1}{2}\)

\(P=\frac{25}{101}\)

a) \(A=\frac{1}{1\cdot3\cdot5}+\frac{1}{3\cdot5\cdot7}+...+\frac{1}{25\cdot27\cdot29}\)

   \(\Rightarrow4A=\frac{4}{1\cdot3\cdot5}+\frac{4}{3\cdot5\cdot7}+...+\frac{4}{25\cdot27\cdot29}\)

\(\Rightarrow4A=\frac{1}{1\cdot3}-\frac{1}{3\cdot5}+\frac{1}{3\cdot5}-\frac{1}{5\cdot7}+...+\frac{1}{25\cdot27}-\frac{1}{27\cdot29}\)

\(\Rightarrow4A=\frac{1}{1\cdot3}-\frac{1}{27\cdot29}=\frac{1}{3}-\frac{1}{783}=\frac{261}{783}-\frac{1}{783}=\frac{260}{783}\)

\(\Rightarrow A=\frac{\frac{260}{783}}{4}=\frac{65}{783}\)

b) \(\left(\frac{1}{1\cdot101}+\frac{1}{2\cdot102}+...+\frac{1}{10\cdot110}\right)x=\frac{1}{1\cdot11}+\frac{1}{2\cdot12}+...+\frac{1}{100\cdot110}\)

\(\Rightarrow100\cdot\left(\frac{1}{1\cdot101}+\frac{1}{2\cdot102}+...+\frac{1}{10\cdot110}\right)x=100\cdot\left(\frac{1}{1\cdot11}+\frac{1}{2\cdot12}+...+\frac{1}{100\cdot110}\right)\)

\(\Rightarrow\left(\frac{100}{1\cdot101}+\frac{100}{2\cdot102}+...+\frac{100}{10\cdot110}\right)x=10\cdot\left(\frac{10}{1\cdot11}+\frac{10}{2\cdot12}+...+\frac{10}{100\cdot110}\right)\)

\(\Rightarrow\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110}\right)x=10\cdot\left(1-\frac{1}{10}+\frac{1}{2}-\frac{1}{12}+...+\frac{1}{100}-\frac{1}{110}\right)\)

\(\Rightarrow\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110}\right)x=10\cdot\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110}\right)\)

\(\Rightarrow x=10\cdot\)

đề có sai ko nhỉ ???

Ta có : \(\frac{1}{10.9}-\frac{1}{9.8}-.....-\frac{1}{2.1}\)

\(=\frac{1}{90}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{9.8}\right)\)

\(=\frac{1}{90}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{8}-\frac{1}{9}\right)\)

\(=\frac{1}{90}-\left(1-\frac{1}{9}\right)\)

\(=\frac{1}{90}-\frac{8}{9}=\frac{-79}{90}\)

\(\frac{1}{1.3.5}+\frac{1}{3.5.7}+\frac{1}{5.7.9}+...+\frac{1}{99.101.103}\)

=\(\frac{1}{4}\left(\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}+...+\frac{4}{99.101.103}\right)\)

=\(\frac{1}{4}\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{99.101}-\frac{1}{101.103}\right)\)

=\(\frac{1}{4}\left(\frac{1}{1.3}-\frac{1}{101.103}\right)\)

=\(\frac{1}{4}.\frac{10406}{31209}\)

=\(\frac{5230}{62418}\)

tui chịu thôi

Ta có: 1/1.3.5 = (1/1.3 - 1/3.5).1/4

          1/3.5.7 = (1/3.5 - 1/5.7).1/4

\(\Rightarrow\) 1/1.3.5 + 1/3.5.7 + 1/5.7.9 + ... + 1/99.101.103 = 1/4.(1/1.3 - 1/3.5 + 1/3.5 - 1/5.7 + ... + 1/99.101 - 1/101.103)

                                                                           = 1/4.(1/3 - 1/10403)

                                                                           = 2600/31209

Tớ nghĩ vậy, nếu đúng thì cho mk biết nha

khoan đã bạn chép nhầm đề rồi thì phải số 1 kia không có dấu gì à?

Đặt \(A=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{99\cdot101}\)

\(2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{99}-\frac{1}{101}\)

\(2A=\frac{1}{3}-\frac{1}{101}=\frac{98}{303}\)

\(\Rightarrow A=\frac{98}{303}\div2=\frac{49}{303}\)

đụ cha mi

mi trù ta thi rớt HK II mà ta giúp mày hả

mấy bài này cũng dễ ẹt nữa

đừng có mơ ta sẽ giúp mày

ha ha ha ha ha ha ha ha ha ha ha ha ha ha ha ha
 

\(B=\left(1+\frac{1}{1\cdot3}\right)\left(1+\frac{1}{2\cdot4}\right)\left(1+\frac{1}{3\cdot5}\right)...\left(1+\frac{1}{99\cdot101}\right)\)

\(B=\frac{2^2}{1\cdot3}\cdot\frac{3^2}{2\cdot4}\cdot\frac{4^2}{3\cdot5}\cdot\cdot\cdot\frac{100^2}{99\cdot101}\)

\(B=\frac{2^2\cdot3^2\cdot4^2\cdot\cdot\cdot100^2}{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot\cdot\cdot99\cdot101}\)

\(B=\frac{\left(2\cdot3\cdot4\cdot\cdot\cdot100\right)\cdot\left(2\cdot3\cdot4\cdot\cdot\cdot100\right)}{\left(1\cdot2\cdot3\cdot\cdot\cdot99\right)\cdot\left(3\cdot4\cdot5\cdot\cdot\cdot101\right)}\)

\(B=\frac{100\cdot2}{1\cdot101}\)

\(B=\frac{200}{101}\)

\(\frac{4}{1\cdot3\cdot5}+\frac{4}{3\cdot5\cdot7}+\frac{4}{5\cdot7\cdot9}+\frac{4}{7\cdot9\cdot11}+\frac{4}{9\cdot11\cdot13}\)

\(=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{9.11}-\frac{1}{11.13}\)

\(=\frac{1}{1.3}-\frac{1}{11.13}\)

\(=\frac{1}{3}-\frac{1}{143}\)

\(=\frac{140}{429}\)