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S=1+2+2^2+2^3+...+2^2002
2S=2+2^2+2^3+2^4+...+2^2003
2S-S=(2+2^2+2^3+2^4+...+2^2003)-(1+2+2^2+2^3+...+2^2002)
2S-S=(2-2)+(2^2-2^2)+(2^3-2^3)+...+(2^2002-2^2002)+(2^2003-1)
S=0+0+0+...+0+(2^2003-1)
S=2^2003-1 (rút gọn)
tick nha
S=1+2+2^2+2^3+...+2^2002
=>2S=2+22+23+24+...+22013
=>2S-S=2+22+23+24+...+22013-1-2-22-23-...-22012
=>S=22013-1
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S=30+32+34+36+...+3200
6S=32+34+36+...+3202
6S-S=(32+34+36+...+3202)-(1+32+34+...+3200)
5S=1+(32-32)+(34-34)+...+(3200-3200)+3202
S=(3200+1):5\(\frac{ }{ }\)
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\(\text{C=1+2-3-4+5+6-7-8+9+...+2002-2003-2004+2005+2006}\)
\(\text{C=1+(2-3-4+5)+(6-7-8+9)+...+(2002-2003-2004+2005)+2006}\)
\(\text{C=1+0+0+...+0+2006}\)
\(\text{C=1+2006}\)
\(C=2007\)
HỌC TỐT!!!
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\(a.\dfrac{3}{2}+\dfrac{-1}{3}< \dfrac{x}{6}< \dfrac{1}{9}+\dfrac{31}{18}\)
\(\Leftrightarrow\dfrac{7}{6}< \dfrac{x}{6}< \dfrac{11}{6}\)
\(\Leftrightarrow7< x< 11\)
\(\Leftrightarrow x\in\left\{8;9;10\right\}\)
\(b.\dfrac{-5}{12}+\dfrac{7}{12}+\dfrac{-1}{12}< \dfrac{x}{12}< \dfrac{2}{15}+\dfrac{1}{5}\)
\(\Leftrightarrow\dfrac{1}{12}< \dfrac{x}{12}< \dfrac{1}{3}\)
\(\Leftrightarrow\dfrac{1}{12}< \dfrac{x}{12}< \dfrac{4}{12}\)
\(\Leftrightarrow1< x< 4\)
\(\Leftrightarrow x\in\left\{2;3\right\}\)
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Lời giải:
Xét tử số:
$X=1+2+2^2+2^3+...+2^{2008}$
$2X=2+2^2+2^3+2^4+....+2^{2009}$
$\Rightarrow 2X-X=(2+2^2+2^3+2^4+....+2^{2009})-(1+2+2^2+...+2^{2008})$
$\Rightarrow X=2^{2009}-1$
$\Rightarrow S=\frac{X}{1-2^{2009}}=\frac{2^{2009}-1}{-(2^{2009}-1)}=-1$
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A = 20 + 21 + 22 + 23 +....+ 22002
2A = 21 + 22 + 23 + 24 +....+ 22003
2A - A = ( 21 + 22 + 23 + 24 +....+ 22003 ) - ( 20 + 21 + 22 + 23 + .... + 22002 )
A = 22003 - 1
Hk tốt
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