`a)` `A=1/2 + 1/(2^2) + 1/(2^3) + … + 1/(2^13) + 1/(2^14)` `b)` `(7x – 3)^2012 = ( 3 – 7x)^2010` `c)` `|x-3|-2x=1` `d)` `(4^(x+2) + 4^(x+1) + 4^x)/2

Question

`a)` `A=1/2 + 1/(2^2) + 1/(2^3) + … + 1/(2^13) + 1/(2^14)`
`b)` `(7x – 3)^2012 = ( 3 – 7x)^2010`
`c)` `|x-3|-2x=1`
`d)` `(4^(x+2) + 4^(x+1) + 4^x)/21 = `(3^(2x) + 3^(2x+1) + 3^(2x+2))/31`

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Daisy 2 tuần 2021-11-28T13:43:41+00:00 1 Answers 5 views 0

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    2021-11-28T13:45:05+00:00

    $\begin{array}{l}a)\quad A = \dfrac12 + \dfrac{1}{2^2} + \dfrac{1}{2^3} + \cdots + \dfrac{1}{2^{13}} + \dfrac{1}{2^{14}}\\ \to 2A = 1 + \dfrac12 + \dfrac{1}{2^2} + \dfrac{1}{2^3} + \cdots + \dfrac{1}{2^{13}} + \dfrac{1}{2^{13}}\\ \to 2A – A = \left(1 + \dfrac12 + \dfrac{1}{2^2} + \cdots + \dfrac{1}{2^{13}} + \dfrac{1}{2^{13}}\right) – \left(\dfrac12 + \dfrac{1}{2^2} + \dfrac{1}{2^3} + \cdots + \dfrac{1}{2^{13}} + \dfrac{1}{2^{14}}\right)\\ \to A = 1 – \dfrac{1}{2^{14}} = \dfrac{2^{14}-1}{2^{14}}\\ b)\quad (7x-3)^{2012} = (3-7x)^{2010}\\ \to (7x-3)^{2012} – (7x-3)^{2010} =0\\ \to (7x-3)^{2010}\left[(7x-3)^2 -1\right] =0\\ \to \left[\begin{array}{l}(7x-3)^{2010} =0\\(7x-3)^2 = 1\end{array}\right.\\ \to \left[\begin{array}{l}7x-3=0\\7x-3= 1\\7x – 3 = -1\end{array}\right.\\ \to \left[\begin{array}{l}x = \dfrac37\\x = \dfrac47\\x = \dfrac27\end{array}\right.\\ Vậy\,\,x = \dfrac27;\, x = \dfrac37;\, x = \dfrac47\\ c)\quad |x-3| – 2x = 1\\ \to \left[\begin{array}{l}x – 3 – 2x = 1\qquad (x \geq 3)\\3 -x – 2x = 1\qquad (x < 3)\end{array}\right.\\ \to \left[\begin{array}{l}x = -4\quad (loại)\\x = \dfrac23\quad (nhận)\end{array}\right.\\ Vậy\,\,x = \dfrac23\\ d)\quad \dfrac{4^{x+2}+4^{x+1} +4^x}{21} = \dfrac{3^{2x} + 3^{2x+1} + 3^{2x+3}}{31}\\ \to \dfrac{16.4^{x}+4.4^{x} +4^x}{21} = \dfrac{3^{2x} + 3.3^{2x} + 27.3^{2x}}{31}\\ \to \dfrac{21.4^x}{21} = \dfrac{31.3^{2x}}{31}\\ \to 4^x = 3^{2x}\\ \to 4^x = 9^x\\ \to \left(\dfrac49\right)^x =1\\ \to x = 0\\ \end{array}$

     

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