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sau đây là phần chữa của mình:
\(=\dfrac{1}{2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}\)
\(=\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{10}\)
= \(\dfrac{3}{10}\)
= \(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
= \(\dfrac{1}{2}-\dfrac{1}{10}\)
= \(\dfrac{2}{5}\)
(y - \(\dfrac{1}{2}\)) : \(\left(\dfrac{1}{2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}\right)\)= \(\dfrac{1}{3}\)
(y\(-\dfrac{1}{2}\)): \(\left(\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\)= \(\dfrac{1}{3}\)
\(\left(y-\dfrac{1}{2}\right):\left(\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{10}\right)=\dfrac{1}{3}\)
\(\left(y-\dfrac{1}{2}\right):\dfrac{3}{10}=\dfrac{1}{3}\)
\(\left(y-\dfrac{1}{2}\right)=\dfrac{1}{10}\)
y = \(\dfrac{3}{5}\)
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}\)
\(=\left(\frac{1}{2}+\frac{1}{20}\right)+\left(\frac{1}{6}+\frac{1}{30}\right)+\frac{1}{12}\)
\(=\frac{11}{20}+\frac{1}{5}+\frac{1}{12}\)
\(=\frac{3}{4}+\frac{1}{12}\)
\(=\frac{10}{12}=\frac{5}{6}\)
\(\Rightarrow x+\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}=\dfrac{47}{42}\\ \Rightarrow x+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}=\dfrac{47}{42}\\ \Rightarrow x+1-\dfrac{1}{6}=\dfrac{47}{42}\\ \Rightarrow x=\dfrac{47}{42}-\dfrac{5}{6}=\dfrac{2}{7}\)
ta nhận thấy
1/2=1-1/2
1/6=1/2-1/3
1/12=1/3-1/4
1/20=1/4-1/5
1/30=1/5-1/6
1/42=1/6-1/7
ta có:
1/2+1/6+1/12+1/20+1/30+1/42
=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7
=1-1/7
=6/7
bn tự hiểu nha
Có công thức \(\dfrac{x}{a\left(a+x\right)}=\dfrac{1}{a}-\dfrac{1}{a+x}\) nhé!
Ví dụ: \(\dfrac{2}{2.4}=\dfrac{1}{2}-\dfrac{1}{4}\)
\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}\)
\(=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{7}-\dfrac{1}{8}\)
\(=1-\dfrac{1}{8}=\dfrac{7}{8}\)
Dấu . tức là nhân nhé!
A = \(\dfrac{1}{20}\) + \(\dfrac{1}{30}\) + \(\dfrac{1}{42}\) + \(\dfrac{1}{56}\) + \(\dfrac{1}{72}\) + \(\dfrac{1}{90}\) + \(\dfrac{1}{110}\) + \(\dfrac{1}{132}\)
A = \(\dfrac{1}{4\times5}\) + \(\dfrac{1}{5\times6}\) + \(\dfrac{1}{6\times7}\)+ \(\dfrac{1}{7\times8}\)+\(\dfrac{1}{8\times9}\)+ \(\dfrac{1}{9\times10}\) + \(\dfrac{1}{10\times11}\)+\(\dfrac{1}{11\times12}\)
A = \(\dfrac{1}{4}\)-\(\dfrac{1}{5}\) +\(\dfrac{1}{5}\)-\(\dfrac{1}{6}\) +.....+\(\dfrac{1}{11}\) - \(\dfrac{1}{12}\)
A = \(\dfrac{1}{4}\) - \(\dfrac{1}{12}\)
A = \(\dfrac{1}{6}\)
\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{90}\)
\(=\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+...+\dfrac{1}{9\times10}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=1-\dfrac{1}{10}\)
\(=\dfrac{9}{10}\)
\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+....+\dfrac{1}{90}\\ =\dfrac{1}{2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+.....+\dfrac{1}{9\cdot10}\\ =\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+.....+\dfrac{1}{9}-\dfrac{1}{10}\\ =\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{10}\\ =1-0,1=0,9\)