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A=đã cho.
=>3A=3+3^2+3^3+3^4+...+3^2012+3^2013.
=>3A-A=3^2013-1.
=>2A=3^2013-1.
=>A=\(\frac{3^{2013-1}}{2}\)
=>B-A=3^2013:2-(3^2013-1)/82.
=>B-A=1/2.
Vậy B-A=1/2.
Bài 1)
Ta có:
A = \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+\dfrac{1}{8^2}\)
A < \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}\)
A < \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}\)
A < \(1-\dfrac{1}{8}\) = \(\dfrac{7}{8}\) < 1
Vậy A < 1
Bài 2)
Ta thấy:
\(\dfrac{2011}{2012+2013}< \dfrac{2011}{2012};\dfrac{2012}{2012+2013}< \dfrac{2012}{2013}\)
\(\Rightarrow\) \(\dfrac{2011}{2012+2013}+\dfrac{2012}{2012+2013}< \dfrac{2011}{2012}+\dfrac{2012}{2013}\)
\(\Rightarrow\) \(\dfrac{2011+2012}{2012+2013}< \dfrac{2011}{2012}+\dfrac{2012}{2013}\)
\(\Rightarrow\) A < B
Bài 3)
Ta có:
B = \(\left(1-\dfrac{1}{1}\right)\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right)......\left(1-\dfrac{1}{20}\right)\)
= \(0.\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)......\left(1-\dfrac{1}{20}\right)\)
= 0
Bài 3)
Ta có:
A = \(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+.....+\dfrac{1}{2^{2012}}\)
\(\Rightarrow\) 2A = \(2\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+.....+\dfrac{1}{2^{2012}}\right)\)
\(\Rightarrow\) 2A = \(2+1+\dfrac{1}{2}+\dfrac{1}{2^2}+.....+\dfrac{1}{2^{2011}}\)
\(\Rightarrow\) 2A - A = \(\left(2+1+\dfrac{1}{2}+\dfrac{1}{2^2}+.....+\dfrac{1}{2^{2011}}\right)\)-\(\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+.....+\dfrac{1}{2^{2012}}\right)\)
\(\Rightarrow\) A = 2 - \(\dfrac{1}{2^{2012}}\) = \(\dfrac{2^{2013}-1}{2^{2012}}\)
Bài 5)
\(\pi\) + 5 \(⋮\) \(\pi\) - 2
\(\Leftrightarrow\) \(\pi\) - 2 + 7 \(⋮\) \(\pi\) - 2
\(\Leftrightarrow\) 7 \(⋮\) \(\pi\) - 2 (vì \(\pi\) - 2 \(⋮\) \(\pi\) - 2)
\(\Leftrightarrow\) \(\pi\) - 2 \(\in\) Ư(7)
\(\Leftrightarrow\) \(\pi\) - 2 \(\in\) \(\left\{\pm1;\pm7\right\}\)
\(\Leftrightarrow\) \(\pi\) \(\in\) \(\left\{1;3;-5;9\right\}\)
Ta có : B=\(\frac{17\left(17^{17}+1\right)}{17\left(17^{18}+1\right)}=\frac{17^{18}+17}{17^{19}+17}>A=\frac{17^{18}+1}{17^{19}+1}\)
Vậy A<B.^-^
A=307/324
B=290/307
Giờ bạn lấy 1 trừ đi 2 phân số ấy là ra 2 phân số: 17/324 va 17/327
Vì 17/324>17/327
Nên phân số A>B
k cho mik nha!
Ta có:
\(\frac{1}{-3}< \frac{x}{3}\le0\)
\(\Rightarrow\frac{1}{-3}< \frac{x}{3}\le\frac{0}{3}\)
\(\Rightarrow\frac{-1}{3}< \frac{x}{3}\le\frac{0}{3}\)
\(\Rightarrow-1< x\le0\)
\(\Rightarrow x=0\)
vậy: \(x=0\)
Đổi 1/-3 = -1/3 ; 0 = 0/3
Suy ra -1 < x <_ 0
Suy ra x = 0
Vậy x = 0
\(2012+\frac{2012}{1+2}+\frac{2012}{1+2+3}+.....+\frac{2012}{1+2+3+....+2011}\)
\(=\frac{2012}{\frac{1\left(1+1\right)}{2}}+\frac{2012}{\frac{2\left(2+1\right)}{2}}+\frac{2012}{\frac{3\left(3+1\right)}{2}}+.....+\frac{2012}{\frac{2011\left(2011+1\right)}{2}}\)
\(=\frac{4024}{1.2}+\frac{4024}{2.3}+\frac{4024}{3.4}+.....+\frac{4024}{2011.2012}\)
\(=4024\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2011}-\frac{1}{2012}\right)\)
\(=4024\left(1-\frac{1}{2012}\right)\)
\(=4024.\frac{2011}{2012}\)
\(=4022\)
Gọi d = UCLN (12n+1; 30n+2)
Ta có: 12n+1 chia hết cho d => 5(12+1) chia hết cho d
vừa nãy mk ấn nhầm, xin lỗi nhé
Gọi d = UCLN(12n+1; 30n+2)
Ta có: 12n+1 chia hết cho d => 5.(12n+1) chia hết cho d
30n+2 chia hết cho d => 2.(30n+2) chia hết cho d
Suy ra 5.(12n+1) - 2.(30n+2) chia hết cho d
=> 60n +5 - 60n +4 chia hết cho d
=> 1 chia hết cho d => d=1
Vậy \(\frac{12n+1}{30n+2}\) là phân số tối giản
\(\frac{n+1}{n-2}\)
\(=\frac{n+3-2}{n-2}\)
\(=\frac{n-2+3}{n-2}\)
\(=\frac{n-2}{n-2}+\frac{3}{n-2}\)
Suy ra n - 2 thuộc ước của 3
Ta có Ư( 3 ) = { -1;-3;1;3 }
Do đó
n - 2 = -1
n = -1 + 2
n = 1
n - 2 = -3
n = -3 + 2
n = -1
n - 2 = 1
n = 1 + 2
n = 3
n - 2 = 3
n = 3 + 2
n = 5
Vậy n = 1;-1;3;5
Ta có:\(\frac{n+1}{n-2}=\frac{n-2+3}{n-2}=1+\frac{3}{n-2}\left(n\ne2\right)\)
Đặt \(A=\frac{n+1}{n-2}\)
Để A nguyên thì 3 chia hết cho n-2. Hay \(\left(n-2\right)\inƯ\left(3\right)\)
Vậy Ư (3) là:[1,-1,3,-3]
Do đó ta có bảng sau:
n-2 | -3 | -1 | 1 | 3 |
n | -1 | 1 | 3 | 5 |
Vậy để A nguyên thì n=-1;1;3;5
Ta có: 3A=3+\(^{3^2+3^3+3^4+3^5+...+3^{2012}+3^{2013}}\)
\(\Rightarrow\)3A-A=2A=(\(3+3^2+3^3+3^4+...+3^{2013}\)) - (\(1-3^{ }-3^2-3^3-3^4-...-3^{2012}\))
\(\Rightarrow\)2A=\(3^{2013}-1\)\(\Rightarrow\)A=\(\left(3^{2013}-1\right):2\)\(\Rightarrow\)B-A=(\(^{\left(3^{2013}:2\right)-\left(\left(3^{2013}-1\right):2\right)\Rightarrow}\)
A = 1 + 3 + 32 +...+ 32012
3A = 3 + 32 + 33 +...+ 32013
3A - A = (3 + 32 + 33 +...+ 32013) - (1 + 3 + 32 +...+ 32012)
2A = 32013 - 1
A = \(\frac{3^{2013}-1}{2}\)
=> B - A = \(\frac{3^{2013}}{2}-\frac{3^{2013}-1}{2}=\frac{3^{2013}-\left(3^{2013}-1\right)}{2}=\frac{3^{2013}-3^{2013}+1}{2}=\frac{1}{2}\)