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\(a)\) \(S=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{2187}\)
\(S=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^7}\)
\(3S=3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^6}\)
\(3S-S=\left(3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^6}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^7}\right)\)
\(2S=3+\frac{1}{3^7}\)
\(2S=\frac{3^8+1}{3^7}\)
\(S=\frac{3^8+1}{3^7}.\frac{1}{2}\)
\(S=\frac{3^8+1}{2.3^7}\)
Vậy \(S=\frac{3^8+1}{2.3^7}\)
Chúc bạn học tốt ~
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Câu 1 : \(\frac{10}{3}\cdot x-\frac{1}{2}=\frac{1}{10}\)
\(\Rightarrow\frac{10}{3}\cdot x=\frac{3}{5}\Rightarrow x=\frac{9}{50}\)
Câu 2 : \(\frac{1}{3}\cdot x+\frac{1}{3}\cdot x=\frac{1}{8}\Rightarrow x\cdot\left(\frac{1}{3}+\frac{1}{3}\right)=\frac{1}{8}\)
\(x\cdot\frac{2}{3}=\frac{1}{8}\Rightarrow x=\frac{3}{16}\)
Câu 3 : \(\left[3x-1\right]\left[\frac{1}{2}x-\frac{2}{5}\right]=0\)
\(\Rightarrow3x-1=0\)hoặc \(\frac{1}{2}x-\frac{2}{5}=0\)
( vô lí ) x = 4/5
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\(E=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)....\left(1-\frac{1}{2006}\right)\left(1-\frac{1}{2007}\right)\)
\(E=\frac{1}{2}.\frac{2}{3}....\frac{2005}{2006}.\frac{2006}{2007}\)
\(E=\frac{1.2.3.4...2005.2006}{2.3.4.5....2006.2007}\)
\(E=\frac{1}{2007}\)
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1 / 5 + x = 3 / 7 + 1 / 3
1 / 5 + x = 16 /21
x = 16 / 21 - 1 / 5
x = 59 / 105
x - 1 / 2 = 2 / 3 - 1 / 5
x - 1 / 2 = 7 / 15
x = 7 / 15 + 1 / 2
x = 29 / 30
3 / 5 * x = 2 / 7+ 1 / 4
3 / 5 * x = 15 / 28
x = 15 / 28 : 3 / 5
x = 25 / 28
7 / 8 : x = 1 / 6 * 2 / 3
7 / 8 : x = 1 / 9
x = 7 / 8 : 1 / 9
x = 63 / 8
\(\dfrac{1}{3}+\dfrac{2}{3}:x-\dfrac{1}{8}=\dfrac{1}{2}\\ \Leftrightarrow\dfrac{2}{3}:x=\dfrac{1}{2}+\dfrac{1}{8}-\dfrac{1}{3}\\ \Leftrightarrow\dfrac{2}{3}:x=\dfrac{12+3-8}{24}\\ \Leftrightarrow\dfrac{2}{3}:x=\dfrac{7}{24}\\ \Leftrightarrow x=\dfrac{2}{3}:\dfrac{7}{24}=\dfrac{2}{3}\times\dfrac{24}{7}=\dfrac{16}{7}\)
31+32:x−81=21⇔32:x=21+81−31⇔32:x=2412+3−8⇔32:x=247⇔x=32:247=32×724=716