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B = (1 + 1/2)(1 + 1/3)(1 + 1/4) ...(1 + 1/100)
= \(\frac{3}{2}.\frac{4}{3}.\frac{5}{4}...\frac{101}{100}\)
= \(\frac{3.4.5....101}{2.3.4...100}=\frac{101}{2}\)
C = \(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{1000}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{999}{1000}\)
\(=\frac{1.2.3...999}{2.3.4....1000}=\frac{1}{1000}\)
c)
\(\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+....+\left(1-\frac{1}{42}\right)+\left(1-\frac{1}{56}\right)\)
\(\left(1+1+1+....+1+1\right)+\left(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{6\times7}+\frac{1}{7\times8}\right)\)(Có 7 số 1)
\(7+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)
\(7+1-\frac{1}{8}=\frac{63}{8}\)
Gợi ý 1 bài c) còn d) e) cũng làm như vậy nhé
Chúc bạn học tốt !!!
a, \(5-\left(\frac{a}{b}+\frac{1}{2}\right)=2\frac{1}{3}\) => \(\frac{a}{b}+\frac{1}{2}=5-2\frac{1}{3}\) => \(\frac{a}{b}+\frac{1}{2}=\frac{8}{3}\) => \(\frac{a}{b}=\frac{8}{3}-\frac{1}{2}\) => \(\frac{a}{b}=\frac{13}{6}\)
b, \((\frac{3}{4}+2\frac{1}{2}):\frac{3}{5-3}=\left(\frac{3}{4}+\frac{5}{4}\right):\frac{3}{5}-1=\frac{9}{4}:\frac{-2}{5}=\frac{-45}{8}\)
a, 5-(\(\frac{a}{b}\)+\(\frac{1}{2}\))=2\(\frac{1}{3}\)
<=>5-\(\frac{a}{b}-\frac{1}{2}\)=\(\frac{7}{3}\)
<=>\(\frac{a}{b}=5-\frac{1}{2}-\frac{7}{3}\)
<=>\(\frac{a}{b}=\frac{13}{6}\)
b,(\(\frac{3}{4}\)+2\(\frac{1}{2}\)):\(\frac{3}{5}\)-3
=(\(\frac{3}{4}\)+\(\frac{5}{2}\)).\(\frac{5}{3}\)-3
=\(\frac{23}{4}\).\(\frac{5}{3}\)-3
=\(\frac{115}{12}\)-3
=\(\frac{115-36}{12}\)
=\(\frac{79}{12}\)
\(\left(1^1+2^2+3^3+4^4+...+2022^{2022}\right)\left(8^2-576:3^2\right)\)
\(=\left(1^1+2^2+3^3+4^4+...+2022^{2022}\right)\left(64-576:3^2\right)\)
\(=\left(1^1+2^2+3^3+4^4+...+2022^{2022}\right)\left(64-64\right)\)
\(=\left(1^1+2^2+3^3+4^4+2022^{2022}\right).0\)
\(=0\)
\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2006}}\)
\(\Rightarrow2A=2.\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2006}}\right)\)
\(=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2005}}\)
\(\Rightarrow2A-A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2005}}-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2006}}\right)\)
\(\Rightarrow A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2005}}-1-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{2006}}\)
\(=2-\frac{1}{2^{2006}}\)