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\(\frac{5}{2.4}+\frac{5}{4.6}+\frac{5}{6.8}+...+\frac{5}{48.50}\)
\(=\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{48}-\frac{1}{50}\right)\)
\(=\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{50}\right)\)
\(=\frac{5}{2}.\frac{12}{25}\)
\(=\frac{6}{5}\)
\(=\left(1+3+5+...+99+101\right)-\left(2+4+6+...98+100\right)\)
Thấy từ 1 đến 100 có (101-1)/2+1=51
=> 1+3+5+....+99+100=(1+101)x50/2=2601
Từ 2 đến 100 có (102-2)/2+1=50
=> 2+4+...+98+100=(2+100)X50/2=2550
=> D=2601-2550=51
\(\frac{5}{2.4}+\frac{5}{4.6}+\frac{5}{6.8}+....+\frac{5}{48.50}\)
\(=\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{48}-\frac{1}{50}\right)\)
\(=\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{50}\right)\)
\(=\frac{5}{2}.\frac{12}{25}=\frac{6}{5}\)
\(\frac{5}{2.4}+\frac{5}{4.6}+\frac{5}{6.8}+...+\frac{5}{48.50}\)
\(=\frac{2}{5}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{48.50}\right)\)
\(=\frac{2}{5}.\left(\frac{4-2}{2.4}+\frac{6-4}{4.6}+\frac{8-6}{6.8}+...+\frac{50-48}{48.50}\right)\)
\(=\frac{2}{5}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{48}-\frac{1}{50}\right)\)
\(=\frac{2}{5}.\left(\frac{1}{2}-\frac{1}{50}\right)\)
\(=\frac{2}{5}.\frac{12}{25}\)
\(=\frac{24}{125}\)
= 1/2 . 2 . ( 2/4.6 - 2/6.8 + .......+ 2/2008.2010)
= 1 . (1/4 - 1/6 + 1/6 - 1/8 +.....+ 1/2010 )
= 1 . ( 1/4 - 1/2010)
= 1 . 1003/4020 = 1003/4020
mik nghĩ bạn viết sai đề phải là 4/2*4 chứ không phải là 4*4/2 nều mà bạn sai đề thì phải giải như sau:
ta có A=2*(2/2*4+2/4*6+2/6*8+....2/2008*2010)
A=2*(1/2-1/4+1/4-1/6+1/6-1/8+.....+1/2008-1/2010)
A=2*(1/2-1/2010)
A=2*502/1005
A=1004/1005
Mk cg tính ra kết quả này nhg thấy sai sai nên cg chưa đăng nè
\(B=\frac{3}{4.6}+\frac{3}{6.8}+\frac{3}{8.10}+...+\frac{3}{20.22}\)
\(=\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}+...+\frac{1}{20.22}\)
\(=\left(\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{20}-\frac{1}{22}\right)\)
\(=\frac{1}{4}-\frac{1}{22}\)
\(=\frac{9}{44}\)
1)\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}\)
\(=\frac{100}{101}\)
2)\(=2\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2008.2010}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{4}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{2010}\right)\)
\(=2\times\frac{502}{1005}\)
\(=\frac{1004}{1005}\)
tự làm tiếp nhé
1.= \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
= \(1-\frac{1}{101}\) = \(\frac{100}{101}\)
2.= \(2\cdot\left(\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+...+\frac{2}{2008\cdot2010}\right)\)
= \(2\cdot\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
= \(2\cdot\left(\frac{1}{2}-\frac{1}{2010}\right)\) = \(2\cdot\frac{502}{1005}\) = \(\frac{1004}{1005}\)
\(\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{x\left(x+2\right)}=\frac{1}{10}\left(ĐKXĐ:x\ne0;x\ne-2\right)\)
\(\Rightarrow\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{x\left(x+2\right)}=\frac{2}{10}\).
\(\Rightarrow\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{1}{5}\).
\(\Rightarrow\frac{1}{4}-\frac{1}{x+2}=\frac{1}{5}\).
\(\Rightarrow\frac{5\left(x+2\right)}{20\left(x+2\right)}-\frac{20}{20\left(x+2\right)}=\frac{4\left(x+2\right)}{20\left(x+2\right)}\)..
\(\Rightarrow5\left(x+2\right)-20=4\left(x+2\right)\).
\(\Rightarrow5x+10-20=4x+8\).
\(\Rightarrow5x-4x=8-10+20\).
\(\Rightarrow x=18\)(thỏa mãn ĐKXĐ) (ĐKXĐ : Điều kiện xác định).
Vậy \(x=18\).
\(=\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{48}-\frac{1}{50}\)
\(=\frac{1}{4}-\frac{1}{50}\)
\(=\frac{23}{100}\)
Nếu bn hỏi cái 1/4,1/6 các kiểu ở đâu ra thì nó là phân tích của 2/4.6 đấy
Chúc bn hok tốt
= 2 . ( 1/4 - 1/6 + 1/6 -1/8 +...+ 1/48 -1/50)
= 2. (1/4- 1/50)
= 2. 23/100
= 46/100= 23/50