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\(A=\left(\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\right)-\left(\frac{1}{3}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\right)\)
\(3^2A=3^2\left(\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\right)-3^2\left(\frac{1}{3}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\right)\)
\(9A=\left(1+\frac{1}{3^2}+...+\frac{1}{3^{98}}\right)-\left(3+\frac{1}{3}+...+\frac{1}{3^{97}}\right)\)
\(9A-A=\left(1-\frac{1}{3^{100}}\right)-\left(3-\frac{1}{3^{99}}\right)\)
\(8A=1-3=-2\)
A=\(\frac{-2}{8}=\frac{-1}{4}\)
\(B=4\left|\frac{-1}{4}\right|+\frac{1}{3^{100}}=1+\frac{1}{3^{100}}=1\)
Vậy B=1
1. A = 75(42004 + 42003 +...+ 42 + 4 + 1) + 25
A = 25 . [3 . (42004 + 42003 +...+ 42 + 4 + 1) + 1]
A = 25 . (3 . 42004 + 3 . 42003 +...+ 3 . 42 + 3 . 4 + 3 + 1)
A = 25 . (3 . 42004 + 3 . 42003 +...+ 3 . 42 + 3 . 4 + 4)
A = 25 . 4 . (3 . 42003 + 3 . 42002 +...+ 3 . 4 + 3 + 1)
A =100 . (3 . 42003 + 3 . 42002 +...+ 3 . 4 + 3 + 1) \(⋮\) 100
\(A=\dfrac{-1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...+\dfrac{1}{3^{100}}\)
\(\Rightarrow\dfrac{1}{3}A=\dfrac{-1}{3^2}+\dfrac{1}{3^3}-\dfrac{1}{3^4}+...-\dfrac{1}{3^{100}}+\dfrac{1}{3^{101}}\)
Cộng vế với vế:
\(A+\dfrac{1}{3}A=\dfrac{-1}{3}-\dfrac{1}{3^2}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{100}}-\dfrac{1}{3^{100}}+\dfrac{1}{3^{101}}\)
\(\Rightarrow\dfrac{4}{3}A=\dfrac{-1}{3}+\dfrac{1}{3^{101}}\)
\(\Rightarrow A=\dfrac{1}{4}\left(\dfrac{1}{3^{100}}-1\right)\)
Do \(\dfrac{1}{3^{100}}< \dfrac{1}{3}< 1\Rightarrow A< 0\)
\(\Rightarrow\left|A\right|=-A=-\dfrac{1}{4}\left(\dfrac{1}{3^{100}}-1\right)=\dfrac{1}{4}\left(1-\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow B=4\left|A\right|+\dfrac{1}{3^{100}}=1-\dfrac{1}{3^{100}}+\dfrac{1}{3^{100}}=1\)
a. \(\frac{20^5.5^{10}}{100^5}\)
\(=\frac{20^5.\left(5^2\right)^5}{100^5}\)
\(=\frac{20^5.25^5}{100^5}\)
\(=\frac{500^5}{100^5}\)
\(=\left(\frac{500}{100}\right)^5\)
\(=5^5=3125\)
b. \(\frac{\left(0,9\right)^5}{\left(0,3\right)^6}\)
\(=\frac{\left(0,9\right)^5}{\left(0,3\right)^5.0,3}\)
\(=\left(\frac{0,9}{0,3}\right)^5.\frac{1}{0,3}\)
\(=3^5.\frac{1}{0,3}\)
\(=810\)
c. \(\frac{6^3+3.6^2+3^3}{-13}\)
\(=\frac{\left(3.2\right)^3+3.\left(3.2\right)^2+3^3}{-13}\)
\(=\frac{3^3\left(2^3+2^2+1\right)}{-13}\)
\(=\frac{3^3.13}{-13}\)
\(=\left(-3\right)^3\)
\(=-27\)