( 78125-15625+3125)/21=65625/21=3125đpcm

5xS=1+1/5+1/25+....+1/3125

5xS-S=(1+1/5+1/25+...+1/3125)-(1/5+1/25+1/125+....+1/15625)

4xS=1-1/15625=15624/15625

1xS=15624/15625:4=15624/15625x1/4=15624/62500=3906/15625

1: \(S=1+\dfrac{1}{3}+\dfrac{1}{9}+...+\dfrac{1}{3^9}\)

\(=\left(\dfrac{1}{3}\right)^0+\left(\dfrac{1}{3}\right)^1+...+\left(\dfrac{1}{3}\right)^9\)

u1=1; q=1/3

\(S_9=\dfrac{u1\cdot\left(1-q^9\right)}{1-q}=\dfrac{1\left(1-\left(\dfrac{1}{3}\right)^9\right)}{1-\dfrac{1}{3}}\)

\(=\dfrac{3}{2}\left(1-\dfrac{1}{3^9}\right)\)

2:

\(S=\left(\dfrac{1}{5}\right)^0+\left(\dfrac{1}{5}\right)^1+...+\left(\dfrac{1}{5}\right)^7\)

\(u1=1;q=\dfrac{1}{5}\)

\(S_7=\dfrac{1\cdot\left(1-q^7\right)}{1-q}=\dfrac{1-\left(\dfrac{1}{5}\right)^7}{1-\dfrac{1}{5}}=\dfrac{5}{4}\left(1-\dfrac{1}{5^7}\right)\)

1, Ta có \(\dfrac{\dfrac{1}{3}}{1}=\dfrac{1}{3};\dfrac{\dfrac{1}{9}}{\dfrac{1}{3}}=\dfrac{1}{3};...\)

-> Là cấp số nhân, q = 1/3 

Ta có \(S_9=1.\dfrac{1-\left(\dfrac{1}{3}\right)^9}{1-\left(\dfrac{1}{3}\right)}\approx1,5\)

b, Ta có \(\dfrac{\dfrac{1}{5}}{1}=\dfrac{1}{5};\dfrac{\dfrac{1}{25}}{\dfrac{1}{5}}=\dfrac{1}{5};...\)

-> Là cấp số nhân, q = 1/5 

\(S_7=\dfrac{1-\left(\dfrac{1}{5}\right)^7}{1-\dfrac{1}{5}}\approx1,25\)

\(A=1+\frac{1}{5}+\frac{1}{25}+...+\frac{1}{78125}\)

\(5A=5+1+\frac{1}{5}+\frac{1}{25}+...+\frac{1}{15625}\)

\(\left(5A-A\right)=\left(5+1+\frac{1}{5}+...+\frac{1}{15625}\right)-\left(1+\frac{1}{5}+...+\frac{1}{78125}\right)\)

\(4A=5-\frac{1}{78125}\)

\(A=5-\frac{1}{312500}\)

\(A=1+\dfrac{1}{5}+\dfrac{1}{25}+\dfrac{1}{125}+...+\dfrac{1}{625}+\dfrac{1}{78125}\)

\(=1+\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+...+\dfrac{1}{5^7}\)

\(5A=5+1+\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^6}\)

\(\Leftrightarrow5A-A=5+1+\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^6}-1-\dfrac{1}{5}-\dfrac{1}{5^2}-\dfrac{1}{5^3}-...-\dfrac{1}{5^7}\)

\(\Leftrightarrow4A=5-\dfrac{1}{5^7}\Leftrightarrow A=\dfrac{5-\dfrac{1}{5^7}}{4}=\dfrac{\dfrac{390624}{78125}}{4}=\dfrac{390624}{312500}=\dfrac{97656}{78125}\)

a) ĐKXĐ: \(x\notin\left\{5;-5\right\}\)

b) Ta có: \(A=\dfrac{2x}{x^2-25}+\dfrac{5}{5-x}-\dfrac{1}{x+5}\)

\(=\dfrac{2x}{\left(x-5\right)\left(x+5\right)}-\dfrac{5}{x-5}-\dfrac{1}{x+5}\)

\(=\dfrac{2x}{\left(x-5\right)\left(x+5\right)}-\dfrac{5\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\dfrac{x-5}{\left(x+5\right)\left(x-5\right)}\)

\(=\dfrac{2x-5x-25-x+5}{\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{-4x-20}{\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{-4\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{-4}{x-5}\)

Để A nguyên thì \(-4⋮x-5\)

\(\Leftrightarrow x-5\inƯ\left(-4\right)\)

\(\Leftrightarrow x-5\in\left\{1;-1;2;-2;4;-4\right\}\)

hay \(x\in\left\{6;4;7;3;9;1\right\}\)(nhận)

Vậy: Để A nguyên thì \(x\in\left\{6;4;7;3;9;1\right\}\)

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