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 A = 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729 

A x 3 = 3 x (1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729) 

= 1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243  A x 3 - A

= 1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243 - (1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729) 

= 1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243 - 1/3 - 1/9 - 1/27 - 1/81 - 1/243 - 1/729 

= 1 - 1/729  A x 2

= 728/729 

A = 364/729 

14 tháng 8 2017

$A=\dfrac{2018.2017-1}{2016.2018+2017}$

$=>A={2018.2016+2018-1}{2016.2018+2017}$

$=>A={2018.2016+2017}{2016.2018+2017}$

$=>A=1$

14 tháng 8 2017

\(A=\dfrac{2018.2017-1}{2018.2016+2017}\)

\(A=\dfrac{2018.\left(2016+1\right)-1}{2018.2016+2017}\)

\(A=\dfrac{2018.2016+2018-1}{2018.2016+2017}\)

\(A=\dfrac{2018.2016+2017}{2018.2016+2017}=1\)

\(B=\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}+\dfrac{1}{2187}\)

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

\(\Rightarrow3B=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^6}\)

\(\Rightarrow3B-B=\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^6}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^7}\right)\)

\(\Rightarrow2B=1-\dfrac{1}{3^7}\Rightarrow B=\dfrac{1-\dfrac{1}{2187}}{2}=\dfrac{1093}{2187}\)

Chúc bạn học tốt!!!

3 tháng 7 2017

a, \(\dfrac{27}{8x^3-1}:\dfrac{3}{2x-1}\)

\(=\dfrac{27}{\left(2x-1\right)\left(4x^2+2x+1\right)}.\dfrac{2x-1}{3}\)

\(=\dfrac{9}{4x^2+2x+1}\)

b, \(\dfrac{8x^3+36x^2+54x+27}{2x+3}=\dfrac{\left(2x+3\right)^3}{2x+3}=\left(2x+3\right)^2\)

10 tháng 10 2017

a) (x2-6xy+9y2):(3y-x)

= (x-3y)2:(3y-x)

=(3y-x)2:(3y-x)

= 3y-x

b) (8x3-1):(4x2+2x+1)

=[(2x)3-1]:(4x2+2x+1)

= (2x-1)(4x2+2x+1):(4x2+2x+1)

= 2x-1

10 tháng 10 2017

c) (4x4-9):(2x2-3)

=(2x2-3)(2x2+3):(2x2-3)

=2x2+3

d) (8x3-27):(4x2+6x+9)

=(2x-3)(4x2+6x+9):(4x2+6x+9)

=2x-3

14 tháng 8 2018

a) \(\left(5x-1\right)^6=729\)

\(\Rightarrow\left[{}\begin{matrix}\left(5x-1\right)^6=3^6\\\left(5x-1\right)^6=\left(-3\right)^6\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}5x-1=3\\5x-1=-3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}5x=4\\5x=-2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=-\dfrac{2}{5}\end{matrix}\right.\)

b) \(\dfrac{8}{25}=\dfrac{2^x}{5^{x-1}}\)

\(\Rightarrow\left[{}\begin{matrix}2^x=2^3\\5^{x-1}=5^2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\x-1=2\end{matrix}\right.\)

\(\Rightarrow x=3\)

Vậy x = 3

c) \(\left(\dfrac{1}{16}\right)^x=\left(\dfrac{1}{2}\right)^{10}\)

\(\Rightarrow\left(\dfrac{1}{2}\right)^{3x}=\left(\dfrac{1}{2}\right)^{10}\)

\(\Rightarrow3x=10\)

\(\Rightarrow x=\dfrac{10}{3}\)

d) \(9^x:3^x=3\)

\(\Rightarrow\left(9:3\right)^x=3\)

\(\Rightarrow3^x=3^1\)

\(\Rightarrow x=1\)