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http://olm.vn/hỏi-đáp/question/584545.html chờ xí tui thấy cái tên rồi giải cho bài 2
2E=1+\(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2003}}\)
2E-E=1-\(\frac{1}{2^{2004}}\)
E=\(\frac{1}{2^{2004}}\)
Ủng hộ mk nha
Ta có: \(\frac{x+1}{2014}+\frac{x+2}{2013}+\frac{x+3}{2012}=\frac{x+4}{2011}+\frac{x+5}{2010}+\frac{x+6}{2009}\)
\(\Rightarrow\frac{x+1}{2014}+1+\frac{x+2}{2013}+1+\frac{x+3}{2012}+1=\frac{x+4}{2011}+1+\frac{x+5}{2010}+1+\frac{x+6}{2009}+1\)
\(\Rightarrow\frac{2015+x}{2014}+\frac{2015+x}{2013}+\frac{2015+x}{2012}=\frac{2015+x}{2011}+\frac{2015+x}{2010}+\frac{2015+x}{2009}\)
\(\Rightarrow\left(2015+x\right)\left(\frac{1}{2014}+\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}\right)=0\)
=> 2015 + x = 0
=> x = -2015
A=1-1/4+1/4-1/7+1/7-1/10+....+1/n-1/(n+3)
A=1-1/(n+3)
vì 1/(n+3)lớn hơn 0 nên 1-1/(n+3)<1
=>A<1
(x-1+3)/9=1/y+2
(x+2)/9=1/(y+2)
tích chéo:x.y+2x+2y=5
phân phối ra rồi tìm ước của 5 sau đó lập bảng là ra
\(A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\)
\(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)
\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)
\(A=\frac{1}{2}-\frac{1}{8}\)
\(A=\frac{3}{8}\)
\(A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\)
\(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)
\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)
\(A=\frac{1}{2}-\frac{1}{8}=\frac{3}{8}\)
mình nhé!
Đặt \(B=1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}\)
\(=\left(1+\frac{1}{99}\right)+\left(\frac{1}{3}+\frac{1}{97}\right)+\left(\frac{1}{5}+\frac{1}{95}\right)+...+\left(\frac{1}{49}+\frac{1}{51}\right)\)
\(=\frac{100}{99}+\frac{100}{3\times97}+\frac{100}{5\times95}+...+\frac{100}{49\times51}\)
\(=100\left(\frac{1}{99}+\frac{1}{3\times97}+\frac{1}{5\times95}+...+\frac{1}{49\times51}\right)\)
Đặt \(C=\frac{1}{1\times99}+\frac{1}{3\times97}+\frac{1}{5\times95}+...+\frac{1}{97\times3}+\frac{1}{99\times1}\)
\(=2\left(\frac{1}{99}+\frac{1}{3\times97}+\frac{1}{5\times95}+...+\frac{1}{49\times51}\right)\)
\(A=\frac{B}{6}=\frac{100}{2}=50\)
Vậy \(A=50\)
\(A=\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.2}+\frac{1}{2.15}+\frac{13}{15.4}\)
\(=\frac{1}{7}\left(\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\right)\)
\(=\frac{1}{7}\left(\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{28}\right)\)
\(=\frac{1}{2}.\frac{13}{28}\)
\(=\frac{13}{56}\)
1.
1.2 +2.3 +...+97.98
=1/3.(1.2.3 +2.3.3 +3.4.3 +...+97.98.3)
=1/3.(1.2.3 - 0.1.2+ 2.3.4 -1.2.3 + 3.4.5 -2.3.4 + ... +97.98.99 -96.97.98)
=1/3 . 97.98.99
= 313698
=>1.2 +2.3 +...+97.98-x=16
=>313698-x=16
=> x=313682
4.
\(\left[\left(\frac{36}{x}-x\right):x-x\right]:x-x=-x\)
\(\left[\left(\frac{36}{x}-x\right):x-x\right]:x=-x+x\)
\(\left[\left(\frac{36}{x}-x\right):x-x\right]:x=0\)
\(\left[\left(\frac{36}{x}-x\right):x-x\right]=0\)
\(\left(\frac{36}{x}-x\right):x=x\Rightarrow\frac{36}{x}-x=x^2\)
\(\frac{36}{x}=x^2+x=x\left(x+1\right)\Rightarrow36=x^2\left(x+1\right)\)
Mà Ư(36)={1;2;3;4;6;9;12;18;36}; 9 là số chính phương duy nhất bé hơn 36=> x2 = 9 => x=3
2 câu kia thì đợi một lúc.
\(\frac{x}{2}-\frac{1}{x}=\frac{1}{12}\Rightarrow\frac{x^2}{2}-\frac{x}{x}=\frac{1}{6}\)
\(\Rightarrow\frac{x^2}{2}-1=\frac{1}{6}\Rightarrow\frac{x^2}{2}=\frac{1}{6}+1=\frac{7}{6}\)
\(\Rightarrow x^2=\frac{7}{6}.2=\frac{7}{3}\)
\(\Rightarrow x=1.5275252317\)
\(\text{Mình nhầm :}\)
\(\frac{x}{2}-\frac{1}{x}=\frac{1}{12}\Rightarrow\frac{x^2}{2}-\frac{x}{x}=\frac{1}{6}\)
\(\Rightarrow\frac{x^2}{2}-1=\frac{1}{6}\Rightarrow\frac{x^2}{2}=\frac{1}{6}+1=\frac{7}{6}\)
\(\Rightarrow x^2=\frac{7}{6}.2=\frac{7}{12}\)
\(\Rightarrow x = 0.76376261583\)