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Đề: X=\(\frac{1}{1+2}\)+\(\frac{1}{1+2+3}\)+.......+\(\frac{1}{1+2+3+4+20}\)
X=\(\frac{1}{2.3:2}\)+\(\frac{1}{3.4:2}\)+\(\frac{1}{4.5:2}\)+......+\(\frac{1}{20.21:2}\)
X=\(\frac{2}{2.3}\)+\(\frac{2}{3.4}\)\(\frac{2}{4.5}\)+........+\(\frac{2}{20.21}\)
X=2.(\(\frac{1}{2}\).3+\(\frac{1}{3}\).4+\(\frac{1}{4}\).5+.....+\(\frac{1}{20}\).21)
X=2.(\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+......+\(\frac{1}{20}\)-\(\frac{1}{21}\))
X=2.(\(\frac{1}{2}\)-\(\frac{1}{21}\))
X=2.(\(\frac{21}{42}\)-\(\frac{2}{42}\))
X=2.\(\frac{19}{42}\)
X=\(\frac{19}{21}\)
Mn xem thử đúng ko nha!
Ta có: \(1+2=\frac{2.3}{2}\); \(1+2+3=\frac{3.4}{2}\); .......... ; \(1+2+3+....+20=\frac{20.21}{2}\)
\(\Rightarrow X=\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+.......+\frac{1}{\frac{20.21}{2}}\)
\(=\frac{2}{2.3}+\frac{2}{3.4}+........+\frac{2}{20.21}=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{20.21}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+..........+\frac{1}{20}-\frac{1}{21}\right)=2.\left(\frac{1}{2}-\frac{1}{21}\right)=2.\frac{19}{42}=\frac{19}{21}\)
a) \(\frac{7}{5}.\frac{-31}{125}.\frac{1}{2}.\frac{10}{17}.\frac{-1}{2^3}=\frac{7.\left(-31\right).1.10.\left(-1\right)}{5.2.125.17.2^3}=\frac{31.7}{17.125.2^3}=\frac{217}{17000}\)
b) \(\left(\frac{17}{28}+\frac{18}{29}-\frac{19}{30}-\frac{20}{31}\right).\left(\frac{-5}{12}+\frac{1}{4}+\frac{1}{6}\right)=\left(\frac{17}{28}+\frac{18}{29}-\frac{19}{30}-\frac{20}{31}\right).0=0\)
c) \(\left(\frac{1}{2}+1\right).\left(\frac{1}{3}+1\right).\left(\frac{1}{4}+1\right)...\left(\frac{1}{99}+1\right)=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}...\frac{100}{99}=\frac{3.4.5...100}{2.3.4...99}=\frac{100}{2}=50\)
d) \(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)...\left(\frac{1}{100}-1\right)=\frac{-1}{2}.\frac{-2}{3}.\frac{-3}{4}...\frac{-99}{100}=\frac{-\left(1.2.3..99\right)}{2.3.4...100}=-\frac{1}{100}\)
e) \(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}...\frac{899}{30^2}=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{29.31}{30.30}=\frac{1.3.2.4.3.5...29.31}{2.2.3.3.4.4...30.30}=\frac{\left(1.2.3..29\right).\left(3.4.5...31\right)}{\left(2.3.4...30\right).\left(2.3.4...30\right)}\)
\(=\frac{1.31}{30.2}=\frac{31}{60}\)
\(\left(6+\left(\frac{1}{2}\right)^3-\left|-\frac{1}{2}\right|\right):\frac{3}{12}\)
\(=\left(6+\frac{1}{8}+\frac{1}{2}\right):\frac{1}{4}\)
=\(\frac{53}{8}:\frac{1}{4}\)
\(=\frac{53}{2}\)
a)30/60,-40/60,45/60,48/60
45/60>30/60>-40/60>-48/60
=3/4>1/2>-2/3>-4/5
Ta có: \(1+2+...+n=\frac{n\left(n+1\right)}{2}\) áp dụng vào bài toán ta có
\(1+\frac{1}{2}\left(1+2\right)+\frac{1}{4}\left(1+2+3\right)+...+\frac{1}{20}\left(1+2+...+20\right)\)
\(=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+...+\frac{1}{20}.\frac{20.21}{2}\)
\(=1+\frac{3}{2}+\frac{4}{2}+...+\frac{21}{2}\)
\(=\frac{1}{2}\left(2+3+4+...+20\right)=\frac{1}{2}.\frac{19.22}{2}=\frac{209}{2}\)
Ta có công thức :
1 + 2 + 3 + ... + n = \(\frac{n\left(n+1\right)}{2}\)
Áp dụng vào bài toán ta được :
\(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{20}\left(1+2+3+...+20\right)\)
\(=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+...+\frac{1}{20}.\frac{20.21}{2}\)
\(=1+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{21}{2}\)
\(=\frac{2+3+4+...+21}{2}=\frac{\frac{21.22}{2}-1}{2}=115\)