\(\frac{4}{1X5}\)+ \(\frac{4}{5X9}\)+ \(\frac{4}{9X13}\)+ ......+ \(\frac{4}{21X15}\)+ \(\frac{3}{x}\)= \(\frac{39}{25}\)
K
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11 tháng 3 2020
Đặt \(A=\frac{7}{1\cdot5}+\frac{7}{5\cdot9}+\frac{7}{9\cdot13}+\frac{7}{13\cdot17}+\frac{7}{17\cdot21}\)
\(\frac{4}{7}A=\frac{4}{7}\left(\frac{7}{1\cdot5}+\frac{7}{5\cdot9}+\frac{7}{9\cdot13}+\frac{7}{13\cdot17}+\frac{7}{17\cdot21}\right)\)
\(\frac{4}{7}A=\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+\frac{4}{17\cdot21}\)
\(\frac{4}{7}A=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+\frac{1}{17}-\frac{1}{21}\)
\(\frac{4}{7}A=1-\frac{1}{21}\)
\(\frac{4}{7}A=\frac{20}{21}\)
\(A=\frac{20}{21}:\frac{4}{7}=\frac{20}{21}\cdot\frac{7}{4}=\frac{5}{3}\)
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
11 tháng 8 2018
Đặt biểu thức trên là A
\(\frac{4xA}{7}=\frac{4}{1x5}+\frac{4}{5x9}+\frac{4}{9x13}+\frac{4}{13x17}+\frac{4}{17x21}\)
\(\frac{4xA}{7}=\frac{5-1}{1x5}+\frac{9-5}{5x9}+\frac{13-9}{9x13}+\frac{17-13}{13x17}+\frac{21-17}{17x21}\)
\(\frac{4xA}{7}=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+\frac{1}{17}-\frac{1}{21}\)
\(\frac{4xA}{7}=1-\frac{1}{21}=\frac{20}{21}\Rightarrow A=\frac{20}{21}.\frac{7}{4}=\frac{5}{3}\)
DP
29 tháng 6 2017
\(\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+\frac{3}{4.5}+\frac{3}{5.6}+...+\frac{3}{9.10}+\frac{77}{2.9}+\frac{77}{9.16}+\frac{77}{16.23}+...+\frac{77}{93.100}\)
Gọi \(\left(\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+......+\frac{3}{9.10}\right)\)là \(A\); \(\left(\frac{77}{2.9}+\frac{77}{9.16}+\frac{77}{16.23}+...+\frac{77}{93.100}\right)\)là B . Ta có :
\(A=\frac{3}{1}.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(A=\frac{3}{1}.\left(\frac{1}{1}-\frac{1}{10}\right)\)
\(A=\frac{3}{1}\cdot\frac{9}{10}=\frac{27}{10}\)
\(B=\frac{77}{7}\left(\frac{1}{2}-\frac{1}{9}+\frac{1}{6}-\frac{1}{16}+\frac{1}{16}-\frac{1}{23}+....+\frac{1}{93}-\frac{1}{100}\right)\)
\(B=\frac{77}{7}\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(B=\frac{77}{7}\cdot\frac{49}{100}=\frac{539}{100}\)
\(\Rightarrow\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+\frac{3}{4.5}+...+\frac{3}{9.10}+\frac{77}{2.9}+\frac{77}{9.16}+\frac{77}{16.23}+...+\frac{77}{93.100}=\frac{27}{10}+\frac{539}{100}=\frac{809}{100}\)
\(\frac{4}{3.6}+\frac{4}{6.9}+\frac{4}{9.12}+\frac{4}{12.15}\)
\(=\frac{4}{3}\cdot\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+\frac{1}{12}-\frac{1}{15}\right)\)
\(=\frac{4}{3}\cdot\left(\frac{1}{3}-\frac{1}{15}\right)\)
\(=\frac{4}{3}\cdot\frac{4}{15}=\frac{16}{45}\)
2 tháng 6 2017
Ta có:\(\frac{1}{3x4}+\frac{2}{4x6}+\frac{3}{6x9}+\frac{4}{9x13}+\frac{5}{13x18}\)
\(=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}\)
\(=\frac{1}{3}-\frac{1}{18}=\frac{6}{18}-\frac{1}{18}=\frac{5}{18}\)
2 tháng 6 2017
đặt \(A=\frac{1}{3.4}+\frac{2}{4.6}+\frac{3}{6.9}+\frac{4}{9.13}+\frac{5}{13.18}\)=\(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}=\frac{1}{3}-\frac{1}{18}=\frac{5}{18}\)
=>A=5/18
vậy........
PL
0
24 tháng 6 2017
\(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}+\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}-\frac{4}{625}}\)
\(=\frac{1\left(\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\right)}{4.\left(\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\right)}+\frac{3.\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}{4.\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}\)
\(=\frac{1}{4}+\frac{3}{4}=1\)
24 tháng 6 2017
\(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}+\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}-\frac{4}{625}}\)
\(=\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{4\left(\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\right)}+\frac{3\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}{4\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}\)
\(=\frac{1}{4}+\frac{3}{4}\)
=1
DP
16 tháng 8 2017
\(4\cdot\left(x-12\right)=2x+164\)
\(\Leftrightarrow4x-48=2x+164\)
\(\Leftrightarrow4x-2x=164+48\)
\(\Leftrightarrow2x=212\Rightarrow x=106\)
Sửa đề : \(\frac{1}{1.5}+\frac{1}{5.9}+...+\frac{1}{401.405}+x=2\frac{1}{405}\)
\(\Leftrightarrow\frac{1}{4}\cdot\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{401}-\frac{1}{405}\right)+x=\frac{811}{405}\)
\(\Leftrightarrow\frac{1}{4}\cdot\left(1-\frac{1}{405}\right)+x=\frac{811}{405}\)
\(\Leftrightarrow\frac{1}{4}\cdot\frac{404}{405}+x=\frac{811}{405}\)
\(\Leftrightarrow\frac{101}{405}+x=\frac{811}{405}\Rightarrow x=\frac{142}{81}\)
16 tháng 4 2020
\(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}+\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}-\frac{4}{625}}\)
\(=\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{4\left(\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\right)}+\frac{3\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}{4\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}\)
\(=\frac{1}{4}+\frac{3}{4}=\frac{4}{4}=1\)
Đề\(\Rightarrow\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{15}-\frac{1}{21}+\frac{3}{x}=\frac{39}{25}\)
\(\Rightarrow1-\frac{1}{21}+\frac{3}{x}=\frac{39}{25}\Rightarrow\frac{20}{21}+\frac{3}{x}=\frac{39}{25}\)
\(\Rightarrow\frac{3}{x}=\frac{39}{25}-\frac{20}{21}\)
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