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2x x 16 = 128
2x = 128 : 16
2 x = 8
2x = 23
3x : 9 = 27
3x = 27 x 9
3x =243
3x = 35
[ 2x + 1 ]3 = 27
2x3 + 13 = 27
2x3 +1 = 27
2x3 = 27 - 1
2x3 = 26
\(x^4\cdot x^7\cdot...\cdot x^{100}\)
\(=x^{4+7+...+100}\)
\(=x^{52\cdot33}=x^{1716}\)
\(x^1\cdot x^2\cdot x^3\cdot...\cdot x^{2006}\)
Ta có : \(x^1\cdot x^2=x^{1+2}=x^3\)
Tương tự : \(x^1\cdot x^2\cdot x^3=x^{1+2+3}=x^6\)
Áp dụng vào bài toán :
\(x^1\cdot x^2\cdot x^3\cdot...\cdot x^{2006}=x^{1+2+3+...+2006}\)
\(\Rightarrow x^{1+2+3+...+2006}=x^{2013021}\)
\(2E=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{59}}.\)
\(E=2E-E=1-\frac{1}{2^{60}}\)
\(3^{x-1}+5.3^{x-1}=162\)
\(\Rightarrow3^{x-1}\left(5+1\right)=162\)
\(\Rightarrow3^{x-1}.6=162\)
\(\Rightarrow3^{x-1}=27\)
\(\Rightarrow3^{x-1}=3^3\)
\(\Rightarrow x-1=3\)
\(\Rightarrow x=4\)
\(3^{x-1}+5.3^{x-1}=162\Rightarrow3^{x-1}\left(1+5\right)=162\Rightarrow3^{x-1}.6=162\)
\(\Rightarrow3^{x-1}=162:6\Rightarrow3^{x-1}=27\Rightarrow x-1=3\Rightarrow x=4\)
\(3+3^2+3^3+...+3^{2012}\)
\(=\left(3+3^2+3^3+3^4\right)+...+\left(3^{2009}+3^{2010}+3^{2011}+3^{2012}\right)\)
\(=3\left(1+3+3^2+3^3\right)+...+3^{2009}\left(1+3+3^2+3^3\right)\)
\(=40\left(3+...+3^{2009}\right)⋮40\)
\(S=2+2^2+2^3+.....+2^{100}\)
\(2S=2^2+2^3+2^4+.....+2^{101}\)
\(2S-S=\left(2^2+2^3+2^4+.....+2^{101}\right)-\left(2+2^2+2^3+....+2^{100}\right)\)
\(S=2^{101}-2\)
36 nha bn
36 nha bạn