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Đáp án + Giải thích các bước giải:
7/1×5+7/5×9+7/9×13+7/13×17+7/17×21
=7/4×(4/1×5+4/5×9+4/9×13+4/13×17+4/17×21)
=7/4×(1−1/5+1/5−1/9+1/9−1/13+1/13−1/17+1/17−1/21)
=7/4×(1+0+0+0+0−1/21)
=7/4×(1−1/21)
=7/4×2021
=5/3
nhớ tick nha
\(\dfrac{7}{1\times5}+\dfrac{7}{5\times9}+\dfrac{7}{9\times13}+\dfrac{7}{13\times17}\)\(+\) \(\dfrac{7}{17\times21}\)
= \(\dfrac{7}{4}\times\)( \(\dfrac{4}{1\times5}\)\(+\) \(\dfrac{4}{5\times9}\)\(+\) \(\dfrac{4}{9\times13}\)\(+\)\(\dfrac{4}{13\times17}\)\(+\)\(\dfrac{4}{17\times21}\))
= 7 \(\times\)(\(\dfrac{1}{1}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{17}-\dfrac{1}{21}\))
= 7\(\times\)(\(\dfrac{1}{1}-\dfrac{1}{21}\))
= \(\dfrac{7}{4}\)\(\times\) \(\dfrac{20}{21}\)
= \(\dfrac{5}{3}\)
\(\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+\frac{3}{4.5}+\frac{3}{5.6}+...+\frac{3}{9.10}+\frac{77}{2.9}+\frac{77}{9.16}+\frac{77}{16.23}+...+\frac{77}{93.100}\)
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 :
\(A=\frac{3}{1}.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(A=\frac{3}{1}.\left(\frac{1}{1}-\frac{1}{10}\right)\)
\(A=\frac{3}{1}\cdot\frac{9}{10}=\frac{27}{10}\)
\(B=\frac{77}{7}\left(\frac{1}{2}-\frac{1}{9}+\frac{1}{6}-\frac{1}{16}+\frac{1}{16}-\frac{1}{23}+....+\frac{1}{93}-\frac{1}{100}\right)\)
\(B=\frac{77}{7}\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(B=\frac{77}{7}\cdot\frac{49}{100}=\frac{539}{100}\)
\(\Rightarrow\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+\frac{3}{4.5}+...+\frac{3}{9.10}+\frac{77}{2.9}+\frac{77}{9.16}+\frac{77}{16.23}+...+\frac{77}{93.100}=\frac{27}{10}+\frac{539}{100}=\frac{809}{100}\)
\(\frac{4}{3.6}+\frac{4}{6.9}+\frac{4}{9.12}+\frac{4}{12.15}\)
\(=\frac{4}{3}\cdot\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+\frac{1}{12}-\frac{1}{15}\right)\)
\(=\frac{4}{3}\cdot\left(\frac{1}{3}-\frac{1}{15}\right)\)
\(=\frac{4}{3}\cdot\frac{4}{15}=\frac{16}{45}\)
3/(1×5) + 3/(5×9) + 3/(9×13) + 3/(13×17) + 3/(17×21)
= 3/4 × (1 - 1/5 + 1/5 - 1/9 + 1/9 - 1/13 + 1/13 - 1/17 + 1/17 - 1/21)
= 3/4 × (1 - 1/21)
= 3/4 × 20/21
= 5/7
\(\frac{4}{1\times5}+\frac{4}{5\times9}+\frac{4}{9\times13}+\frac{4}{13\times17}+\frac{4}{17\times21}\)\(=\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+\frac{1}{17}-\frac{1}{21}\)\(=1-\frac{1}{21}=\frac{20}{21}\)
#Y/n
4/5x9 + 4/9x13 + 4/13x17 + 4/ 17x21 + 4/ 21x25=27.911628241
Ta có:
\(\frac{4}{5\times9}+\frac{4}{9\times13}+\frac{4}{13\times17}+\frac{4}{17\times21}+\frac{4}{21\times25}\)
= \(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+\frac{1}{17}-\frac{1}{21}+\frac{1}{21}-\frac{1}{25}\)
= \(\frac{1}{5}-\left(\frac{1}{9}+\frac{1}{9}\right)-\left(\frac{1}{13}-\frac{1}{13}\right)-\left(\frac{1}{17}-\frac{1}{17}\right)-\left(\frac{1}{21}-\frac{1}{21}\right)-\frac{1}{25}\)
= \(\frac{1}{5}-\frac{1}{25}\)
= \(\frac{5}{25}-\frac{1}{25}=\frac{4}{25}\)
Ta có: \(1\frac{4}{5}+2\frac{5}{7}+3\frac{4}{5}+4\frac{5}{7}\)
\(=\left(1\frac{4}{5}+3\frac{4}{5}\right)+\left(2\frac{5}{7}+4\frac{5}{7}\right)\)
\(=\left(\frac{9}{5}+\frac{19}{5}\right)+\left(\frac{19}{7}+\frac{33}{7}\right)\)
\(=\frac{28}{5}+\frac{52}{7}=13\frac{1}{35}\)
= ( \(1\frac{4}{5}\)+ \(3\frac{4}{5}\)) + ( \(2\frac{5}{7}\)+ \(4\frac{5}{7}\))
= \(4\frac{4}{5}\) + \(6\frac{5}{7}\)
= \(\frac{24}{5}\) + \(\frac{47}{7}\)
= ...... ( tính nốt nhé )
Đặt \(A=\frac{7}{1\cdot5}+\frac{7}{5\cdot9}+\frac{7}{9\cdot13}+\frac{7}{13\cdot17}+\frac{7}{17\cdot21}\)
\(\frac{4}{7}A=\frac{4}{7}\left(\frac{7}{1\cdot5}+\frac{7}{5\cdot9}+\frac{7}{9\cdot13}+\frac{7}{13\cdot17}+\frac{7}{17\cdot21}\right)\)
\(\frac{4}{7}A=\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+\frac{4}{17\cdot21}\)
\(\frac{4}{7}A=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+\frac{1}{17}-\frac{1}{21}\)
\(\frac{4}{7}A=1-\frac{1}{21}\)
\(\frac{4}{7}A=\frac{20}{21}\)
\(A=\frac{20}{21}:\frac{4}{7}=\frac{20}{21}\cdot\frac{7}{4}=\frac{5}{3}\)
Đặt biểu thức trên là A
\(\frac{4xA}{7}=\frac{4}{1x5}+\frac{4}{5x9}+\frac{4}{9x13}+\frac{4}{13x17}+\frac{4}{17x21}\)
\(\frac{4xA}{7}=\frac{5-1}{1x5}+\frac{9-5}{5x9}+\frac{13-9}{9x13}+\frac{17-13}{13x17}+\frac{21-17}{17x21}\)
\(\frac{4xA}{7}=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+\frac{1}{17}-\frac{1}{21}\)
\(\frac{4xA}{7}=1-\frac{1}{21}=\frac{20}{21}\Rightarrow A=\frac{20}{21}.\frac{7}{4}=\frac{5}{3}\)