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#It's the moment when you're in good mood, you accidentally click back =.=
1) Calculate
\(P=1\frac{1}{3}.1\frac{1}{8}.1\frac{1}{15}....1\frac{1}{63}.1\frac{1}{80}\)
\(=\frac{4}{3}.\frac{9}{8}.\frac{16}{15}....\frac{64}{63}.\frac{81}{80}\)
\(=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}....\frac{8.8}{7.9}.\frac{9.9}{8.10}\)
\(=\frac{2.9}{10}=\frac{9}{5}\)
ta có: 10010 + 1 > 10010 - 1
⇒ A = \(\frac{100^{10}+1}{100^{10}-1}< \frac{100^{10}+1-2}{100^{10}-1-2}=\frac{100^{10}-1}{100^{10}-3}=B\)
vậy A < B
\(A=\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(\frac{A}{7}=\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(\frac{A}{7}=\frac{7-2}{2.7}+\frac{11-7}{7.11}+\frac{14-11}{11.4}+\frac{15-14}{14.15}+\frac{28-15}{15.28}\)
\(\frac{A}{7}=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}=\frac{1}{2}-\frac{1}{28}=\frac{13}{28}\)
\(A=7.\frac{13}{28}\)
\(A=\frac{13}{4}\)
Ta có :
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2009}}+\frac{1}{2^{2010}}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2008}}+\frac{1}{2^{2009}}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2008}}+\frac{1}{2^{2009}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2009}}+\frac{1}{2^{2010}}\right)\)
\(A=1-\frac{1}{2^{2010}}\)
\(A=\frac{2^{2010}-1}{2^{2010}}\)
Vậy \(A=\frac{2^{2010}-1}{2^{2010}}\)
Chúc bạn học tốt
Bạn giải cũng được đấy alibaba nguyễn, nhưng theo mình thì làm cách này dễ hiểu hơn!
Ta có: \(C=\frac{\frac{2010}{1}+\frac{2009}{2}+\frac{2008}{3}+...+\frac{1}{2010}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}}\)
Đặt \(A=\frac{2010}{1}+\frac{2009}{2}+\frac{2008}{3}+...+\frac{1}{2010}\)
\(A=\frac{2010}{1}+1+\frac{2009}{1}+1+\frac{2008}{1}+1+...+\frac{1}{2010}+1-2010\)
\(=\frac{2011}{1}+\frac{2011}{2}+\frac{2011}{3}+...+\frac{2011}{2010}-\frac{2011.2010}{2011}\)
\(=2011\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2010}-\frac{2010}{2011}\right)\)
Đặt \(B=\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}\)
\(B=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}-1\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2010}-\frac{2010}{2011}\)
Ta có: \(C=\frac{A}{B}=2011\)(lấy A-B)
Ta có :
\(2010A=\dfrac{2010^{2012}+2010}{2010^{2012}+1}=\dfrac{2010^{2012}+1+2009}{2010^{2012}+1}=1+\dfrac{2009}{2010^{2012}+1}\)
\(2010B=\dfrac{2010^{2011}+2010}{2010^{2011}+1}=\dfrac{2010^{2011}+1+2009}{2010^{2011}+1}=1+\dfrac{2009}{2010^{2011}+1}\)
Vì \(1+\dfrac{2009}{2010^{2012}+1}< 1+\dfrac{2009}{2010^{2011}+1}\Rightarrow A< B\)
~ Học tốt ~
\(C=\frac{\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}}{\frac{5}{2008}-\frac{5}{2009}-\frac{5}{2010}}+\frac{\frac{2}{2007}-\frac{2}{2008}-\frac{2}{2009}}{\frac{3}{2007}-\frac{3}{2008}-\frac{3}{2009}}\)
\(=\frac{\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}}{5.\left(\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}\right)}+\frac{2.\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)}{3.\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)}\)
\(=\frac{1}{5}+\frac{2}{3}\)
\(=\frac{13}{15}\)
a) A= 1/2010+1+2/2009+1+3/2008+1+...+2009/2+1+1
= 2011/2010+20011/2009+2011/2008+...+2011/2+2011/2011
= 2011(1/2+1/3+1/4+...+1/2011)
Ta có: B= 1/2+1/3+1/4+...+1/2011
suy ra A/B= 2011
a) Đặt A= \(\frac{1+2+2^2+2^3+...+2^{2009}}{1-2^{2010}}\)
Đặt S = 1 + 2 + 22 + 23 + ... + 22009
=> 2S = 2 + 22 + 23 + ... + 22010
=> 2S - S = (2 + 22 + 23 + ... + 22010) - (1 + 2 + 22 + 23 + .. + 22009)
=> S = 22010 - 1
=> S = - 1 - 22010
\(\Rightarrow A=\frac{-1-2^{2010}}{1-2^{2010}}=-1\)
Vậy \(\frac{1+2+2^2+2^3+...+2^{2009}}{1-2^{2010}}=-1\)
b) Đặt: \(A=\frac{1}{299.297}-\frac{1}{297.295}-\frac{1}{295.293}-...-\frac{1}{3.1}\)
\(\Rightarrow-2A=-\frac{2}{299.297}+\frac{2}{297.295}+\frac{2}{295.293}+...+\frac{2}{3.1}\)
\(\Rightarrow-2A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{295.297}-\frac{2}{297.299}\)
\(\Rightarrow-2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{295}-\frac{1}{297}-\frac{1}{297.299}\)
\(\Rightarrow-2A=1-\frac{1}{297}-\frac{2}{88803}\)
\(\Rightarrow-2A=\frac{296}{297}-\frac{2}{88803}=\frac{88504}{88803}-\frac{2}{88803}=\frac{88502}{88803}\)
\(\Rightarrow A=\frac{88502}{88803}:\left(-2\right)=\frac{44251}{88803}\)
Vậy \(\frac{1}{299.297}-\frac{1}{297.295}-\frac{1}{295.293}-...-\frac{1}{3.1}=\frac{44251}{88803}\)
c) Đặt \(B=\frac{12}{1.3.5}+\frac{12}{3.5.7}+\frac{12}{5.7.9}+...+\frac{12}{25.27.29}\)
\(\Rightarrow\frac{B}{3}=\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}+...+\frac{12}{25.27.29}\)
\(\Rightarrow\frac{B}{3}=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+\frac{1}{5.7}-\frac{1}{7.9}+...+\frac{1}{25.27}-\frac{1}{27.29}\)
\(\Rightarrow\frac{B}{3}=\frac{1}{1.3}-\frac{1}{27.29}\)
\(\Rightarrow\frac{B}{3}=\frac{1}{3}-\frac{1}{783}=\frac{261}{783}-\frac{1}{783}=\frac{260}{783}\)
\(\Rightarrow B=\frac{260}{783}.3=\frac{260}{261}\)
Vậy \(\frac{12}{1.3.5}+\frac{12}{3.5.7}+\frac{12}{5.7.9}+...+\frac{12}{25.27.29}=\frac{260}{261}\)
Duyệt mk nha!!!