Cho biểu thức :
\(A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2014}}\)
Hãy so sánh A với \(\dfrac{3}{2}\)
K
Khách
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28 tháng 2 2021
Bạn thiếu đề rồi phải là trừ hay cộng j j chứ.
Xét:
`A+B=2+1/2+1/3+1/4+......+1/4026+1/3+1/5+1/7+......+1/4025`
`1/2+1/3+1/4+......+1/4026+1/3+1/5+1/7+......+1/4025>0`
`=>A+B>2`
Mà `1 2013/2014<2`
`=>A+B>1 2013/2014`
20 tháng 8 2021
2)Ta có: \(2^{332}< 2^{333}=\left(2^3\right)^{111}=8^{111}\)
\(3^{223}>3^{222}=\left(3^2\right)^{111}=9^{111}\)
Vì \(8^{111}< 9^{111}\) mà \(2^{332}< 8^{111},3^{223}>9^{111}\) nên suy ra \(2^{332}< 3^{223}\)
Vậy \(2^{332}< 3^{223}\)
20 tháng 8 2021
1) \(A=\dfrac{10^{2013}+1}{10^{2014}+1}\Rightarrow10A=\dfrac{10^{2014}+10}{10^{2014}+1}=\dfrac{10^{2014}+1}{10^{2014}+1}+\dfrac{9}{10^{2014}+1}=1+\dfrac{9}{10^{2014}+1}\)
\(B=\dfrac{10^{2014}+1}{10^{2015}+1}\Rightarrow10B=\dfrac{10^{2015}+10}{10^{2015}+1}=\dfrac{10^{2015}+1}{10^{2015}+1}+\dfrac{9}{10^{2015}+1}=1+\dfrac{9}{10^{2015}+1}\)Vì: \(10^{2014}+1< 10^{2015}+1\Rightarrow\dfrac{9}{10^{2014}+1}>\dfrac{9}{10^{2015}+1}\Rightarrow1+\dfrac{9}{10^{2014}+1}>1+\dfrac{9}{10^{2015}+1}\)
Nên suy ra \(10A>10B\Rightarrow A>B\)
AH
Akai Haruma
Giáo viên
15 tháng 7 2023
Lời giải:
$M=\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+...+\frac{2014}{5^{2014}}$
$5M=1+\frac{2}{5}+\frac{3}{5^2}+...+\frac{2014}{5^{2013}}$
$\Rightarrow 4M=5M-M=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2013}}-\frac{2014}{5^{2014}}$
$4M+\frac{2014}{5^{2014}}=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2013}}$
$5(4M+\frac{2014}{5^{2014}})=5+1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2012}}$
$\Rightarrow 4(4M+\frac{2014}{5^{2014}})=5-\frac{1}{5^{2013}}$
$M=\frac{5}{16}-\frac{1}{16.5^{2013}-\frac{2014}{4.5^{2014}}$
17 tháng 8 2021
\(A=-\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)...\left(1-\dfrac{1}{2014^2}\right)\)
\(A=\dfrac{\left(1\cdot3\right)\left(2\cdot4\right)\left(3\cdot5\right)...\left(2012\cdot2014\right)\left(2013\cdot2015\right)}{\left(2\cdot2\right)\left(3\cdot3\right)\left(4\cdot4\right)...\left(2013\cdot2013\right)\left(2014\cdot2014\right)}\)
\(A=\dfrac{\left(1\cdot2\cdot3\cdot...\cdot2012\cdot2013\right)\left(3\cdot4\cdot5\cdot...\cdot2014\cdot2015\right)}{\left(2\cdot3\cdot4\cdot...\cdot2013\cdot2014\right)\left(2\cdot3\cdot4\cdot...\cdot2013\cdot2014\right)}\)
\(A=\dfrac{1\cdot2015}{2014\cdot2}=\dfrac{2015}{4028}\)
Vì \(\dfrac{2015}{4028}>-\dfrac{1}{2}\) nên A > B
2 tháng 4 2017
A=\(\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot\cdot\cdot\dfrac{-2015}{2016}\)
=\(-\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\cdot\cdot\dfrac{2015}{2016}\)
=\(\dfrac{-1}{2016}>\dfrac{-1}{2015}\)
Vậy\(A>\dfrac{-1}{2015}\)
24 tháng 12 2020
Ta có: \(3\cdot A=1+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\)
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}\)
Do đó:
\(3\cdot A-A=1+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}-\dfrac{1}{3}-\dfrac{1}{3^2}-...-\dfrac{1}{3^{100}}\)
hay \(2\cdot A=1-\dfrac{1}{3^{100}}\)
\(\Leftrightarrow A=\left(1-\dfrac{1}{3^{100}}\right):2\)
\(\Leftrightarrow A=\left(1-\dfrac{1}{3^{100}}\right)\cdot\dfrac{1}{2}\)
\(\Leftrightarrow A=\dfrac{1}{2}-\dfrac{1}{2\cdot3^{100}}< \dfrac{1}{2}\)
hay A<B
MV
29 tháng 4 2017
\(A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2014}}\\ 3A=3+1+\dfrac{1}{3}+...+\dfrac{1}{3^{2013}}\\ 3A-A=\left(3+1+\dfrac{1}{3}+...+\dfrac{1}{3^{2013}}\right)-\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2014}}\right)\\ 2A=3-\dfrac{1}{3^{2014}}\\ A=\left(3-\dfrac{1}{3^{2014}}\right):2\\ A=3:2-\dfrac{1}{3^{2014}}:2\\ A=\dfrac{3}{2}-\dfrac{1}{3^{2014}\cdot2}< \dfrac{3}{2}\)
Vậy \(A< \dfrac{3}{2}\)
3 tháng 3 2022
a: \(P=\dfrac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)
b: Thay x=1/9 vào P, ta được:
\(P=\dfrac{1}{3}:\left(\dfrac{1}{9}+\dfrac{1}{3}+1\right)=\dfrac{1}{3}:\dfrac{1+3+9}{9}=\dfrac{1}{3}\cdot\dfrac{9}{13}=\dfrac{3}{13}\)
\(A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2014}}\)
\(3A=3\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2014}}\right)\)
\(3A=3+1+\dfrac{1}{3}+...+\dfrac{1}{3^{2013}}\)
\(3A-A=\left(3+1+...+\dfrac{1}{3^{2013}}\right)-\left(1+\dfrac{1}{3}+...+\dfrac{1}{3^{2014}}\right)\)
\(2A=3-\dfrac{1}{3^{2014}}\Rightarrow A=\dfrac{3}{2}-\dfrac{\dfrac{1}{3^{2014}}}{2}< \dfrac{3}{2}\)
Vậy \(A< \dfrac{3}{2}\)
A=1+13+132+133+...+132014A=1+13+132+133+...+132014
3A=3(1+13+132+133+...+132014)3A=3(1+13+132+133+...+132014)
3A=3+1+13+...+1320133A=3+1+13+...+132013
3A−A=(3+1+...+132013)−(1+13+...+132014)3A−A=(3+1+...+132013)−(1+13+...+132014)
2A=3−132014⇒A=32−1320142<32