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Giải:
a) Gọi dãy đó là A, ta có:
\(A=\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2014}}\)
\(2A=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2013}}\)
\(2A-A=\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2013}}\right)-\left(\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2014}}\right)\)
\(A=\dfrac{1}{2}-\dfrac{1}{2^{2014}}\)
Vì \(\dfrac{1}{2}< 1;\dfrac{1}{2^{2014}}< 1\) nên \(\dfrac{1}{2}-\dfrac{1}{2^{2014}}< 1\)
\(\Rightarrow A< 1\)
b) \(A=\dfrac{10^{11}-1}{10^{12}-1}\) và \(B=\dfrac{10^{10}+1}{10^{11}+1}\)
Ta có:
\(A=\dfrac{10^{11}-1}{10^{12}-1}\)
\(10A=\dfrac{10^{12}-10}{10^{12}-1}\)
\(10A=\dfrac{10^{12}-1+9}{10^{12}-1}\)
\(10A=1+\dfrac{9}{10^{12}-1}\)
Tương tự:
\(B=\dfrac{10^{10}+1}{10^{11}+1}\)
\(10B=\dfrac{10^{11}+10}{10^{11}+1}\)
\(10B=\dfrac{10^{11}+1+9}{10^{11}+1}\)
\(10B=1+\dfrac{9}{10^{11}+1}\)
Vì \(\dfrac{9}{10^{12}-1}< \dfrac{9}{10^{11}+1}\) nên \(10A< 10B\)
\(\Rightarrow A< B\)
1:
a: Vì \(\dfrac{-4}{3}=\dfrac{-4\cdot3}{3\cdot3}=\dfrac{-12}{9}=\dfrac{12}{9}\\ \Rightarrow\dfrac{-4}{3}=\dfrac{12}{9}\)
b: Vì : \(-2\cdot3=-6\\ -6\cdot8=-48\)
nên 2 p/s ko bằng nhau
a, Chia hết cho 3 thì nhóm 2 số thành 1 cặp ; chia hết cho 7 thì nhóm 3 số thành 1 cặp
b, Đề phải là A = 2009.2011
Có :A = 2009.(2010+1) = 2009.2010+2009
= 2009.2010+2010-1 = 2010.(2009+1)-1 = 2010^2-1
Vì 2010^2-1 < 2010^2 = B => A < B
c, A = (3^3)^150 = 27^150
B = (5^2)^150 = 25^150
Vì 27^150 > 25^150 => A > B
k mk nha
2:
a: A=1+2+2^2+2^3+2^4
=>2A=2+2^2+2^3+2^4+2^5
=>A=2^5-1
=>A=B
b: C=3+3^2+...+3^100
=>3C=3^2+3^3+...+3^101
=>2C=3^101-3
=>\(C=\dfrac{3^{101}-3}{2}\)
=>C=D
Ta có:
\(\left\{\begin{matrix}5^{27}=\left(5^3\right)^9=125^9\\2^{63}=\left(2^7\right)^9=128^9\end{matrix}\right\}\Rightarrow5^{27}< 2^{63}\left(1\right)\)
\(\left\{\begin{matrix}2^{63}=\left(2^9\right)^7=512^7\\5^{28}=\left(5^4\right)^7=625^7\end{matrix}\right\}\Rightarrow2^{63}< 5^{28}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow5^{27}< 2^{63}< 5^{28}\) (đpcm)
a) \(A=1+2+2^2+...+2^{63}\)
\(\Rightarrow2A=2.\left(1+2+2^2+...+2^{63}\right)\)
\(\Rightarrow2A=2+2^2+...+2^{64}\)
\(\Rightarrow2A-A=2+2^2+...+2^{64}-\left(1+2+2^2+...+2^{63}\right)\)
\(\Rightarrow A=2+2^2+...+2^{64}-1-2-2^2-...-2^{63}\)
\(\Rightarrow A=2^{64}-1\)
Vì \(2^{64}-1=2^{64}-1\Rightarrow A=B\)
b) \(A=3^4+3^5+...+3^{20}\)
\(\Rightarrow3A=3^5+3^6+...+3^{21}\)
\(\Rightarrow3A-A=3^5+3^6+...+3^{21}-3^4-3^5-...-3^{20}\)
\(\Rightarrow2A=3^{21}-3^4\)
\(\Rightarrow A=\frac{3^{21}-3^4}{2}\)
Mà \(B=\frac{3^{21}-3^4}{2}\Rightarrow A=B\)