\(\frac{1}{2}+\frac{2}{2}=??\)
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HH
30 tháng 11 2015
\(1+\frac{1+\frac{1+\frac{3}{2}}{2}}{2}=1+\frac{1+\frac{\frac{5}{2}}{2}}{2}=1+\frac{1+\frac{5}{4}}{2}=1+\frac{\frac{9}{4}}{2}=1+\frac{9}{8}=\frac{17}{8}\)
\(1+\frac{2}{1+\frac{2}{1+\frac{2}{3}}}=1+\frac{2}{1+\frac{2}{\frac{5}{3}}}=1+\frac{2}{1+\frac{6}{5}}=1+\frac{2}{\frac{11}{5}}=1+\frac{10}{11}=\frac{21}{11}\)
\(1+\frac{1+\frac{1+\frac{2}{3}}{3}}{3}=1+\frac{1+\frac{\frac{5}{3}}{3}}{3}=1+\frac{1+\frac{5}{9}}{3}=1+\frac{\frac{14}{9}}{3}=1+\frac{14}{27}=\frac{41}{27}\)
\(\frac{3}{\frac{3}{\frac{3}{\frac{3}{2}+1}+1}+1}+1=1+\frac{3}{\frac{3}{\frac{3}{\frac{5}{2}}+1}+1}=1+\frac{3}{\frac{3}{\frac{6}{5}+1}+1}=1+\frac{3}{\frac{15}{11}+1}=\frac{59}{26}\)
suy ra
\(\frac{\frac{17}{18}}{\frac{21}{11}}-x=\frac{187}{378}-x=\frac{\frac{41}{27}}{\frac{59}{26}}=\frac{1066}{1593}\Rightarrow x=-\frac{1297}{7434}\)
8 tháng 9 2015
Xét \(P=\sqrt{\frac{1}{1^2}+\frac{1}{a^2}+\frac{1}{\left(a+1\right)^2}}\) với a>0
\(P^2=1+\frac{1}{a^2}+\frac{1}{\left(a+1\right)^2}\)
\(=\frac{a^2\left(a+1\right)^2+\left(a+1\right)^2+a^2}{a^2\left(a+1\right)^2}\)
\(=\frac{a^2\left(a^2+2a+1+1\right)+\left(a+1\right)^2}{a^2\left(a+1\right)^2}\)
\(=\frac{a^4+2a^2\left(a+1\right)+\left(a+1\right)^2}{a^2\left(a+1\right)^2}\)
\(=\frac{\left(a^2+a+1\right)^2}{a^2\left(a+1\right)^2}\)
\(=\left(\frac{a^2+a+1}{a\left(a+1\right)}\right)^2\)
Do a>o nên \(P=\frac{a^2+a+1}{a\left(a+1\right)}=1+\frac{1}{a}-\frac{1}{a+1}\)
Áp dụng kết quả của P ta có:
\(A=\left(1+\frac{1}{1}-\frac{1}{2}\right)+\left(1+\frac{1}{2}+\frac{1}{3}\right)+....+\left(1+\frac{1}{2012}-\frac{1}{2013}\right)\) \(A=2012+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-.....+\frac{1}{2012}-\frac{1}{2013}\right)\)
\(A=2012+1-\frac{1}{2013}\)
\(A=2013-\frac{1}{2013}=\frac{4052168}{2013}\)
Vậy \(A=\frac{4052168}{2013}\)
S
27 tháng 7 2019
\(2A=1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+.....-\frac{1}{2^{99}}\Rightarrow2A+A=3A=\left(1-\frac{1}{2}+\frac{1}{2^2}-....-\frac{1}{2^{99}}\right)+\left(\frac{1}{2}-\frac{1}{2^2}+......-\frac{1}{2^{100}}\right)=1-\frac{1}{2^{100}}=\frac{2^{100}-1}{2^{100}}\Rightarrow A=\frac{2^{100}-1}{3.2^{100}}\)
\(2,4B=2+\frac{1}{2}+\frac{1}{2^3}+.....+\frac{1}{2^{97}}\Rightarrow4B-B=3B=\left(2+\frac{1}{2}+....+\frac{1}{2^{97}}\right)-\left(\frac{1}{2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\right)=2-\frac{1}{2^{99}}=\frac{2^{100}-1}{2^{99}}\Rightarrow B=\frac{2^{100}-1}{3.2^{99}}\)
\(3,C=\frac{1}{2}-\frac{1}{2^4}+\frac{1}{2^7}-....-\frac{1}{2^{58}}\Rightarrow8C=4-\frac{1}{2}+\frac{1}{2^4}-.....-\frac{1}{2^{55}}\Rightarrow8C+C=9C=\left(4-\frac{1}{2}+\frac{1}{2^4}-....-\frac{1}{2^{55}}\right)+\left(\frac{1}{2}-\frac{1}{2^4}+\frac{1}{2^7}-....-\frac{1}{2^{58}}\right)=4-\frac{1}{2^{58}}=\frac{2^{60}-1}{2^{58}}\Rightarrow C=\frac{2^{60}-1}{9.2^{58}}\)
15 tháng 4 2018
\(b)\) Đặt \(B=\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}\) ta có :
\(B>\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}=\frac{3+3+3+3+3}{15}=\frac{3.5}{15}=\frac{15}{15}=1\)
\(\Rightarrow\)\(B>1\) \(\left(1\right)\)
Lại có :
\(B< \frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}=\frac{3+3+3+3+3}{10}=\frac{3.5}{10}=\frac{15}{10}< \frac{20}{10}=2\)
\(\Rightarrow\)\(B< 2\) \(\left(2\right)\)
Từ (1) và (2) suy ra :
\(1< B< 2\) ( đpcm )
Vậy \(1< B< 2\)
Chúc bạn học tốt ~
\(\frac{1}{2}+\frac{2}{2}=\frac{3}{2}=1\frac{1}{2}\)
Học Tốt !!!!
\(\frac{3}{2}\)
nha các bạn ^.^