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\(2A=1+\frac{1}{2}+...+\frac{1}{2^{49}}\)
\(2A-A=1-\frac{1}{2^{50}}\)
\(A=1-\frac{1}{2^{50}}\)=> A bé hơn 1
tương tự nha
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\)
\(2A=2.\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\right)\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{48}}+\frac{1}{2^{49}}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{48}}+\frac{1}{2^{49}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\right)\)
\(A=1-\frac{1}{2^{50}}< 1\)
1.
a.
\(\frac{1}{3}+\left(\frac{1}{5}-\frac{1}{7}\right)\)
\(=\frac{1}{3}+\frac{1}{5}-\frac{1}{7}\)
\(=\frac{35-21-15}{105}\)
\(=-\frac{1}{105}\)
b.
\(\frac{3}{5}-\left(\frac{3}{4}-\frac{1}{2}\right)\)
\(=\frac{3}{5}-\frac{3}{4}+\frac{1}{2}\)
\(=\frac{12-15+10}{20}\)
\(=\frac{7}{20}\)
c.
\(\frac{4}{7}-\left(\frac{2}{5}+\frac{1}{3}\right)\)
\(=\frac{4}{7}-\frac{2}{5}-\frac{1}{3}\)
\(=\frac{60-42-35}{105}\)
\(=-\frac{17}{105}\)
2.
a.
\(S=-\frac{1}{1\times2}-\frac{1}{2\times3}-\frac{1}{3\times4}-...-\frac{1}{\left(n-1\right)\times n}\)
\(S=-\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{\left(n-1\right)\times n}\right)\)
\(S=-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\right)\)
\(S=-\left(1-\frac{1}{n}\right)\)
\(S=-1+\frac{1}{n}\)
b.
\(S=-\frac{4}{1\times5}-\frac{4}{5\times9}-\frac{4}{9\times13}-...-\frac{4}{\left(n-4\right)\times n}\)
\(S=-\left(\frac{4}{1\times5}+\frac{4}{5\times9}+\frac{4}{9\times13}+...+\frac{4}{\left(n-4\right)\times n}\right)\)
\(S=-\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{n-4}-\frac{1}{n}\right)\)
\(S=-\left(1-\frac{1}{n}\right)\)
\(S=-1+\frac{1}{n}\)
Chúc bạn học tốt
a) \(x:\left(-\frac{1}{3}\right)^3=-\frac{1}{3}\)
\(\Rightarrow x=\left(-\frac{1}{3}\right).\left(-\frac{1}{3}\right)^3\)
\(\Rightarrow x=\left(-\frac{1}{3}\right)^4\)
\(\Rightarrow x=\frac{1}{81}\)
Vậy \(x=\frac{1}{81}.\)
b) \(\frac{3}{4}:\frac{41}{99}=x:\frac{75}{90}\)
\(\Rightarrow\frac{297}{164}=x:\frac{75}{90}\)
\(\Rightarrow x=\frac{297}{164}.\frac{75}{90}\)
\(\Rightarrow x=\frac{495}{328}\)
Vậy \(x=\frac{495}{328}.\)
c) \(x+\left|-\frac{1}{2}\right|=3\frac{1}{3}-4\frac{1}{2}\)
\(\Rightarrow x+\frac{1}{2}=\frac{10}{3}-\frac{9}{2}\)
\(\Rightarrow x+\frac{1}{2}=-\frac{7}{6}\)
\(\Rightarrow x=\left(-\frac{7}{6}\right)-\frac{1}{2}\)
\(\Rightarrow x=-\frac{5}{3}\)
Vậy \(x=-\frac{5}{3}.\)
Chúc bạn học tốt!
Với mọi \(x\in Z\) ta có:
\(1+2+3+..+n=\frac{n\left(n+1\right)}{2}\)
=> \(\frac{1}{1+2+3+..+n}=\frac{2}{n\left(n+1\right)}=2\left[\frac{1}{n\left(n+1\right)}\right]=2\left(\frac{1}{n}-\frac{1}{n+1}\right)\)
Có:
\(\frac{1}{1+2}=2\left(\frac{1}{2}-\frac{1}{3}\right)\)
\(\frac{1}{1+2+3}=2\left(\frac{1}{3}-\frac{1}{4}\right)\)
.......................................................
\(\frac{1}{1+2+3+4+...+99}=2\left(\frac{1}{99}-\frac{1}{100}\right)\)
Nên:
\(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+4+..+99}+\frac{1}{50}\)
\(=2\left(\frac{1}{2}-\frac{1}{3}\right)+2\left(\frac{1}{3}-\frac{1}{4}\right)+...+2\left(\frac{1}{99}-\frac{1}{100}\right)+\frac{1}{50}\)
\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)+\frac{1}{50}\)
\(=2\left(\frac{1}{2}-\frac{1}{100}\right)+\frac{1}{50}=2\cdot\frac{49}{100}+\frac{1}{50}=\frac{49}{50}+\frac{1}{50}=1\)
Cảm ơn bạn (chị ) nhiều !
Công nhận chị học giỏi thật đấy !