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a, 2A = 2+2^2+....+2^2012
A=2A-A=(2+2^2+....+2^2012)-(1+2+2^2+....+2^2011) = 2^2012-1 > 2^2011-1 = B
=> A>B
b, A = 2009.(2010+1) = 2009.2010+2009 = (2009.2010+2010)-1 = 2010.(2009+1)-1 = 2010.2010-1 = 2010^2-1 < 2010^2 = B
=> A<B
c, A = (10^3)^10 = 1000^10 < 1024^10 = (2^10)^10 = 2^100 = B
=> A<B
k mk nha
\(\frac{-2}{5}.\left(\frac{5}{17}-\frac{9}{15}\right)-\frac{-2}{5}.\left(\frac{2}{17}+\frac{-2}{5}\right)\)
\(=\frac{-2}{5}.\frac{5}{17}-\frac{-2}{5}.\frac{3}{5}-\frac{-2}{5}.\frac{2}{17}-\frac{-2}{5}.\frac{-2}{5}\)
\(=\frac{-2}{5}.\left(\frac{5}{17}-\frac{2}{17}\right)-\frac{-2}{5}.\left(\frac{3}{5}+\frac{-2}{5}\right)\)
\(=\frac{-2}{5}.\frac{3}{17}-\frac{-2}{5}.\frac{1}{5}\)
\(=\frac{-2}{5}.\left(\frac{3}{17}-\frac{1}{5}\right)\)
\(=\frac{-2}{5}.\frac{-2}{85}\)
\(=\frac{4}{425}\)
\(\frac{-2}{5}.\left(\frac{5}{17}-\frac{9}{15}\right)-\frac{-2}{5}.\left(\frac{2}{17}+\frac{-2}{5}\right)\)
= \(\frac{-2}{5}.\frac{-26}{85}-\frac{-2}{5}.\frac{-24}{85}\)
= \(\frac{-2}{5}.\left(\frac{-26}{85}-\frac{-24}{85}\right)\)
= \(\frac{-2}{5}.\frac{-2}{85}\)
= \(\frac{4}{425}\)
\(\left(\frac{1}{x}-\frac{2}{3}\right)^2-\frac{1}{16}=0\)
\(\Leftrightarrow\left(\frac{1}{x}-\frac{2}{3}\right)^2-\left(\frac{1}{4}\right)^2=0\)
\(\Leftrightarrow\left(\frac{1}{x}-\frac{2}{3}+\frac{1}{4}\right)\left(\frac{1}{x}-\frac{2}{3}-\frac{1}{4}\right)=0\)
\(\Leftrightarrow\left(\frac{1}{x}-\frac{5}{12}\right)\left(\frac{1}{x}-\frac{11}{12}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{1}{x}-\frac{5}{12}=0\\\frac{1}{x}-\frac{11}{12}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{1}{x}=\frac{5}{12}\\\frac{1}{x}=\frac{11}{12}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{12}{11}\\x=\frac{12}{5}\end{cases}}\)
Vậy....
\(\left(\frac{1}{x}-\frac{2}{3}\right)^2-\frac{1}{16}=0\)
\(\Rightarrow\left(\frac{1}{x}-\frac{2}{3}\right)^2=\frac{1}{16}\)
\(\Rightarrow\left(\frac{1}{x}-\frac{2}{3}\right)^2=\left(\frac{1}{4}\right)^2\)
\(\Rightarrow\frac{1}{x}-\frac{2}{3}=\frac{1}{4}\)
\(\Rightarrow\frac{1}{x}=\frac{11}{12}\)
\(\Rightarrow x=\frac{11}{12}\)
Ta có: 3C= \(2+\frac{2}{3}+\frac{2}{3^2}+\frac{2}{3^3}+...+\frac{2}{3^{49}}\) suy ra \(3C-C=2C=2-\frac{2}{3^{50}}\Rightarrow C=1-\frac{1}{3^{50}}=\frac{3^{50}-1}{3^{50}}\)
\(\frac{101+100+99+98+...+3+2+1}{101-100+99-98+...+3-2+1}\)
\(=\frac{\frac{101.102}{2}}{51}\)
\(=101\)
(x+1/2).(2/3-2x)=0
=> x+1/2=0 hoặc 2/3-2x=0
+) x+1/2=0 +) 2/3-2x=0
X= - 1/2 2x=2/3
x=1/3
Vậy x ...............
a) ta có: \(M=1+3+3^2+3^3+...+3^{119}\)
\(M=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^{117}+3^{118}+3^{119}\right)\)
\(M=\left(1+3+3^2\right)+3^3.\left(1+3+3^2\right)+...+3^{117}.\left(1+3+3^2\right)\)
\(M=\left(1+3+3^2\right).\left(1+3^3+...+3^{117}\right)\)
\(M=13.\left(1+3^3+...+3^{117}\right)⋮13\left(đpcm\right)\)
b) ta có: \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};...;\frac{1}{2010^2}< \frac{1}{2009.2010}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2010^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2009.2010}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2009}-\frac{1}{2010}\)
\(=1-\frac{1}{2010}< 1\)
\(\Rightarrow N=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2010^2}< 1\left(đpcm\right)\)
a, \(M=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^{117}+3^{118}+3^{119}\right)\)
\(=\left(1+3+3^2\right)+3^3\left(1+3+3^2\right)+...+3^{117}\left(1+3+3^2\right)\)
\(=\left(1+3+3^2\right)\left(1+3^3+3^6+...+3^{117}\right)\)
\(=13.\left(1+3^3+...+3^{117}\right)⋮13\)
b, \(N=\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{2010.2010}\)
\(< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2009.2010}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2009}-\frac{1}{2010}\)
\(=1-\frac{1}{2010}=\frac{2009}{2010}< 1\)
\(\Rightarrow N< 1\)
\(\Rightarrow2A=2+2^2+...+2^{1011}\)
\(\Rightarrow2A-A=\left(2+2^2+..+2^{1011}\right)-\left(1+2+...+2^{2010}\right)\)
\(\Rightarrow A=2^{1011}-1\)
Ta có: \(A=2^0+2^1+2^2+...+2^{2010}\)
\(\Rightarrow2A=2^1+2^2+2^3+...+2^{2010}+2^{2011}\)
\(\Rightarrow2A-A=\left(2^1+2^2+2^3+...+2^{2010}+2^{2011}\right)-\left(2^0+2^1+2^2+...+2^{2010}\right)\)
\(\Rightarrow A=2^{2011}-1\)