Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)
\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)
\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)
\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)
\(A=1+3^2+3^4+...+3^{102}\)
\(9A=3^2+3^4+...+3^{102}+3^{104}\)
\(\Rightarrow9A-A=3^{104}-1\)
\(\Rightarrow8A=3^{104}-1\)
\(\Rightarrow A=\dfrac{3^{104}-1}{8}\)
`Answer:`
`\frac{x-1}{34}+\frac{x-2}{34}=\frac{x-3}{32}+\frac{x-4}{31}`
`<=>\frac{x-1}{34}-1+\frac{x-2}{33}-1=\frac{x-3}{32}-1+\frac{x-4}{31}-1`
`<=>\frac{x-35}{34}+\frac{x-35}{33}=\frac{x-35}{32}+\frac{x-35}{31}`
`<=>(x-35)(\frac{1}{34}+\frac{1}{33}-\frac{1}{32}-\frac{1}{31})=0`
`<=>x-35=0`
`<=>x=35`
\(D=1+3+3^2+3^3+3^4+...+3^{2022}\)
\(3D=3.\left(1+3+3^2+3^3+3^4+...+3^{2022}\right)\)
\(3D=3+3^2+3^3+3^4+3^5+...+3^{2023}\)
\(3D-D=\left(3+3^2+3^3+3^4+3^5+...+3^{2023}\right)-\left(1+3+3^2+3^3+3^4+...+3^{2022}\right)\)
\(2D=\left(3^{2023}-1\right)\)
\(D=\left(3^{2023}-1\right):2\)
3D=3+3^2+...+3^2023
=>2D=3^2023-1
=>\(D=\dfrac{3^{2023}-1}{2}\)
a) =\(\left(\frac{1}{4}\right)^3\cdot2^5=\frac{1}{2^6}2^5=0,5\)
b) \(=\left(\frac{1}{8}\right)^3\cdot8^4\cdot10^4=80.000\)
c) \(=8^2\cdot\frac{4^5}{2^{20}}=\frac{2^6\cdot2^{10}}{2^{20}}=\frac{1}{2^4}=0,0625\)
d) = \(\frac{81^{11}\cdot3^{17}}{27^{10}\cdot9^{15}}=\frac{3^{44}\cdot3^{17}}{3^{30}\cdot3^{30}}=\frac{3^{61}}{3^{60}}=3\)
Ta có: ( 17 + 34 + a ):3=32
suy ra : 17 + 34 + a = 96
a = 96 -(17 + 34)
a = 45
Vậy a = 45
Ta có (17+34+a):3=32
=>17+34+a=32.3=96
=>a=96-(17+34)
=>a=45
Vậy a=45
a) \(A=2+2^2+2^3+...+2^{2017}\)
\(2A=2^2+2^3+2^4+...+2^{2018}\)
\(2A-A=\left(2^2+2^3+2^4+...+2^{2018}\right)-\left(2+2^2+2^3+...+2^{2017}\right)\)
\(A=2^{2018}-2\)
b) \(C=1+3^2+3^4+...+3^{2018}\)
\(3^2\cdot C=3^2+3^4+3^6+...+3^{2020}\)
\(9C-C=\left(3^2+3^4+3^6+...+3^{2020}\right)-\left(1+3^2+3^4+...+3^{2018}\right)\)
\(8C=3^{2020}-1\)
\(\Rightarrow C=\dfrac{3^{2020}-1}{8}\)
\(Toru\)