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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
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) \(\left(\frac{1}{1\cdot101}+\frac{1}{2\cdot102}+...+\frac{1}{10\cdot110}\right)x=\frac{1}{1\cdot11}+\frac{1}{2\cdot12}+...+\frac{1}{100\cdot110}\)
\(\Rightarrow100\cdot\left(\frac{1}{1\cdot101}+\frac{1}{2\cdot102}+...+\frac{1}{10\cdot110}\right)x=100\cdot\left(\frac{1}{1\cdot11}+\frac{1}{2\cdot12}+...+\frac{1}{100\cdot110}\right)\)
\(\Rightarrow\left(\frac{100}{1\cdot101}+\frac{100}{2\cdot102}+...+\frac{100}{10\cdot110}\right)x=10\cdot\left(\frac{10}{1\cdot11}+\frac{10}{2\cdot12}+...+\frac{10}{100\cdot110}\right)\)
\(\Rightarrow\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110}\right)x=10\cdot\left(1-\frac{1}{10}+\frac{1}{2}-\frac{1}{12}+...+\frac{1}{100}-\frac{1}{110}\right)\)
\(\Rightarrow\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110}\right)x=10\cdot\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110}\right)\)
\(\Rightarrow x=10\cdot\)
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
\(B=\left(1+\frac{1}{1\cdot3}\right)\left(1+\frac{1}{2\cdot4}\right)\left(1+\frac{1}{3\cdot5}\right)...\left(1+\frac{1}{99\cdot101}\right)\)
\(B=\frac{2^2}{1\cdot3}\cdot\frac{3^2}{2\cdot4}\cdot\frac{4^2}{3\cdot5}\cdot\cdot\cdot\frac{100^2}{99\cdot101}\)
\(B=\frac{2^2\cdot3^2\cdot4^2\cdot\cdot\cdot100^2}{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot\cdot\cdot99\cdot101}\)
\(B=\frac{\left(2\cdot3\cdot4\cdot\cdot\cdot100\right)\cdot\left(2\cdot3\cdot4\cdot\cdot\cdot100\right)}{\left(1\cdot2\cdot3\cdot\cdot\cdot99\right)\cdot\left(3\cdot4\cdot5\cdot\cdot\cdot101\right)}\)
\(B=\frac{100\cdot2}{1\cdot101}\)
\(B=\frac{200}{101}\)
\(\frac{4}{1\cdot3\cdot5}+\frac{4}{3\cdot5\cdot7}+\frac{4}{5\cdot7\cdot9}+\frac{4}{7\cdot9\cdot11}+\frac{4}{9\cdot11\cdot13}\)
\(=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{9.11}-\frac{1}{11.13}\)
\(=\frac{1}{1.3}-\frac{1}{11.13}\)
\(=\frac{1}{3}-\frac{1}{143}\)
\(=\frac{140}{429}\)
\(\frac{2.6.10+6.10.14+10.14.18+...+194.198.202}{1.3.5+3.5.7+...+97.99.101}\)
\(=\frac{2^3.1.3.5+2^3.3.5.7+2^3.97.99.101}{1.3.5+3.5.7+...+97.99.101}\)
\(=\frac{2^3\left(1.3.5+3.5.7+...+97.99.101\right)}{1.3.5+3.5.7+...+97.99.101}\)
\(=\frac{2^3}{1}=8\)
Vậy A = 8
#)Giải :
a)\(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{24.25}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{24}-\frac{1}{25}\)
\(=\frac{1}{5}-\frac{1}{25}\)
\(=\frac{4}{25}\)
b)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}\)
\(=\frac{100}{101}\)
a) \(\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{24.25}\)
= \(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{24}-\frac{1}{25}\)
= \(\frac{1}{5}-\frac{1}{25}\)
= \(\frac{4}{25}\)
b) \(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\)
= \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)
= \(1-\frac{1}{101}\)
= \(\frac{100}{101}\)
c) \(5\frac{2}{7}.\frac{8}{11}+5\frac{2}{7}.\frac{5}{11}-5\frac{2}{7}.\frac{2}{11}\)
= \(5\frac{2}{7}.\left(\frac{8}{11}+\frac{5}{11}-\frac{2}{11}\right)\)
= \(5\frac{2}{7}\)
= \(\frac{37}{7}\)
\(A=\frac{1}{2.3}+\frac{1}{6.5}+\frac{1}{10.7}+\frac{1}{14.9}+...+\frac{1}{198.101}\)
\(=\frac{2}{2.6}+\frac{2}{6.10}+\frac{2}{10.14}+\frac{2}{14.18}+...+\frac{2}{198.202}\)
\(=\frac{1}{2}.\left(\frac{4}{2.6}+\frac{4}{6.10}+\frac{4}{10.14}+\frac{4}{14.18}+...+\frac{4}{198.202}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{6}+\frac{1}{6}-\frac{1}{10}+\frac{1}{10}-\frac{1}{14}+\frac{1}{14}-\frac{1}{18}+...+\frac{1}{198}-\frac{1}{202}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{202}\right)\)
\(=\frac{1}{2}.\frac{50}{101}=\frac{25}{101}\)
Ai đúng mình k