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a) Đặt A = 1 + 2 + 22 + ... + 22008 (1)
=> 2A = 2 + 22 + 23 + ... + 22009 (2)
Lấy (2) trừ (1) theo vế ta có :
2A - A = (2 + 22 + 23 + ... + 22009) - (1 + 2 + 22 + ... + 22008)
A = 22009 - 1
Khi đó B = \(\frac{2^{2009}-1}{1-2^{2009}}=\frac{2^{2009}-1}{-\left(2^{2009}-1\right)}=-1\)
b) Ta có A = \(\frac{20^{10}+1}{20^{10}-1}\)
=> A - 1 = \(\frac{20^{10}+1-20^{10}+1}{20^{10}}=\frac{2}{20^{10}}\)
Lại có B = \(\frac{20^{10}-1}{20^{10}-3}\)
=> B - 1 = \(\frac{20^{10}-1-20^{10}+3}{20^{10}-3}=\frac{2}{2^{10}-3}\)
Vì \(\frac{2}{2^{10}}< \frac{2}{2^{10}-3}\)
=> A - 1 < B - 1
=> A < B
a) \(B=\frac{1+2+2^2+2^3+...+2^{2008}}{1-2^{2009}}\)
Đặt \(Q=1+2+2^2+...+2^{2008}\)
\(2Q=2+2^2+2^3+...+2^{2009}\)
\(2Q-Q=2+2^2+2^3+...+2^{2009}-1-2-2^2-...-2^{2008}\)
\(\Rightarrow Q=2^{2009}-1\)
Ta thấy \(Q\) là số đối của \(2^{2009}-1\)
\(\Rightarrow B=-1\)
Vậy \(B=-1\).
b) Ta có: \(A=\frac{20^{10}+1}{20^{10}-1}=\frac{20^{10}-1+2}{20^{10}-1}=1+\frac{2}{20^{10}-1}\)
Ta lại có: \(B=\frac{20^{10}-1}{20^{10}-3}=\frac{20^{10}-3+2}{20^{10}-3}=1+\frac{2}{20^{10}-3}\)
Vì \(\frac{2}{20^{10}-1}< \frac{2}{20^{10}-3}\) nên \(1+\frac{2}{20^{10}-1}< 1+\frac{2}{20^{10}-3}\)
\(\Rightarrow A< B\)
Vậy \(A< B\).
\(2\dfrac{1}{3}.3=\dfrac{7}{3}.3=7.\\ \left(\dfrac{2}{5}-\dfrac{3}{4}\right)-\dfrac{2}{5}=\dfrac{2}{5}-\dfrac{3}{4}-\dfrac{2}{5}=-\dfrac{3}{4}.\\ \dfrac{-10}{11}.\dfrac{4}{7}+\dfrac{-10}{11}.\dfrac{3}{7}+1\dfrac{10}{11}.\\ =\dfrac{-10}{11}\left(\dfrac{4}{7}+\dfrac{3}{7}-1\right).\\ =\dfrac{-10}{11}.\left(1-1\right)=0.\)
1) 2\(\dfrac{1}{3}\).3=\(\dfrac{7}{3}\).3=7.
2) (2/5 -3/4) -2/5 = 2/5 -3/4 -2/5 = -3/4.
3) \(\dfrac{-10}{11}.\dfrac{4}{7}+\dfrac{-10}{11}.\dfrac{3}{7}+1\dfrac{10}{11}=\dfrac{1}{11}\left(-\dfrac{40}{7}-\dfrac{30}{7}+21\right)=\dfrac{1}{11}.\left(-10+21\right)=1\).
\(A=\dfrac{1}{2}+\dfrac{1}{2}^2+\dfrac{1}{2}^3+...+\dfrac{1}{2}\)10
\(= (\dfrac{1}{2}+\dfrac{1}{2}^9)+(\dfrac{1}{2}^2+\dfrac{1}{2}^8)+(\dfrac{1}{2}^3+\dfrac{1}{2}^7)+\dfrac{1}{2}\)10
\(= \dfrac{257}{512}+\dfrac{65}{256}+\dfrac{17}{128}+\dfrac{1}{1024}\)
\(=( \dfrac{514}{1024}+\dfrac{136}{1024})+(\dfrac{260}{1024}+\dfrac{1}{1024})\)
\(=\dfrac{650}{1024}+\dfrac{261}{1024}\)
\(=\dfrac{911}{1024}\)
\(A=\dfrac{\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{10}}}{\dfrac{2}{2}+\dfrac{2}{2^2}+...+\dfrac{2}{2^{10}}}=\dfrac{\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{10}}}{2\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{10}}\right)}=\dfrac{1}{2}\)
A = \(\dfrac{\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+....+\dfrac{1}{2^{10}}}{\dfrac{2}{2}+\dfrac{2}{2^2}+\dfrac{2}{2^3}+...+\dfrac{2}{2^{10}}}\)
= \(\dfrac{\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}}{2\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\right)}\)
= \(\dfrac{1}{2}\)
12+22+...+102=12+2(1+1)+...+10(9+1)
=1+2+1.2+...+9.10+10=(1+2+...+10)+(1.2+...+9.10)=55+A (Đặt A=1.2+...+9.10)
A=1.2+2.3+...+9.10
=>3A=1.2.3+2.3.3+...+9.10.3
=1.2.3+2.3.(4-1)+...+9.10(11-8)
=1.2.3+2.3.4-1.2.3+...+9.10.11-8.9.10
=9.10.11=990
=>A=990:3=330
=>12+22+...+102=330+55=385