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hỗn số thì tử phải lớn hơn mẫu thì chia tử với mẫu thì mới ra hỗn số
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Sai đâu bỏ qua nhé, hơi to mới lại mk tính máy tính ra : \(\frac{77}{30}\)nên ko chắc nhé
\(2+\frac{1}{1+\frac{1}{1+\frac{1}{3+\frac{1}{4}}}}=2+\frac{1}{1+\frac{1}{1+\frac{1}{\frac{13}{4}}}}\)
\(=2+\frac{1}{1+\frac{1}{1+\frac{4}{13}}}=2+\frac{1}{1+\frac{1}{\frac{17}{3}}}\)
\(=2+\frac{1}{1+\frac{3}{17}}=2+\frac{1}{\frac{20}{17}}=2+\frac{17}{20}=\frac{57}{20}\)
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1 : 29 x ( 19 -13 ) - 19 x ( 29 - 13 )
= 29 x 6 - 19 x 16
= 174 - 304
= - 130
2 : 1 - \(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
= 1 - \(\frac{1}{100}\)
= \(\frac{99}{100}\)
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\(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{20}\left(1+2+3+...+20\right)=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+...+\frac{1}{20}.\frac{20.21}{2}=1+\frac{3}{2}+\frac{4}{2}+...+\frac{21}{2}=1+\frac{24.19}{2}=229\)
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\(E=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{200}\left(1+2+....+200\right)\)
\(=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+....+\frac{1}{200}.\frac{200.201}{2}\)
\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+....+\frac{201}{2}\)
\(=\frac{2+3+4+...+201}{2}\)
\(=\frac{\frac{201.202}{2}-1}{2}=10150\)
\(\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+\dfrac{1}{1+2+3+4}+......+\dfrac{1}{1+2+......+50}\)
\(=\dfrac{1}{\dfrac{2.3}{2}}+\dfrac{1}{\dfrac{3.4}{2}}+\dfrac{1}{\dfrac{4.5}{2}}+......+\dfrac{1}{\dfrac{50.51}{2}}\)
\(=\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+........+\dfrac{2}{50.51}\)
\(=2\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+......+\dfrac{1}{99.100}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{100}\right)\)
\(=2.\dfrac{49}{100}\)
\(=\dfrac{49}{50}\)
Xét thừa số tổng quát: \(1+2+3+...+n=\dfrac{n\left(n+1\right)}{2}\)
Suy ra: \(\dfrac{1}{1+2+3+...+n}=\dfrac{1}{\dfrac{n\left(n+1\right)}{2}}=\dfrac{2}{n\left(n+1\right)}\)
Dễ r:v