\(\dfrac{1}{1.3}\) + \(\dfrac{1}{3.5}\) + \(\dfrac{1}{5.7}\) +......+ \(\dfrac{1}{(2x-1)(2x+1)}\) = \(\dfrac{49}{99}\)
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Để : \(3⋮\left(n+2\right)\)
Vì \(n\in N\Rightarrow\left(n+2\right)\in N\)
Mà : \(3⋮\left(n+2\right)\)
\(\Rightarrow\left(n+2\right)\in\left\{Ư\left(3\right)\right\}\)
Ta có : \(Ư\left(3\right)=\left\{\pm1;\pm3\right\}\)
Do đó ta có bảng :
n+2 : 1 -1 3 -3
n : -1 -3 1 -5
Vậy.........
\(2^{x+1}-1=63\\ 2^{x+1}=64\\ 2^{x+1}=2^6\\ =>X+1=6\\ =>x=5\)
\(\left(X-3\right)\cdot4^5=4^8\\ X-3=64\\ =>X=67\)
Sửa đề
\(\dfrac{2}{1^2}\cdot\dfrac{6}{2^2}\cdot\dfrac{12}{3^3}\cdot.......\cdot\dfrac{110}{10^2}\cdot x=-20\)
\(\dfrac{2}{1\cdot1}\cdot\dfrac{2\cdot3}{2\cdot2}\cdot\cdot\cdot\cdot\dfrac{11\cdot10}{10\cdot10}\cdot x=-20\)
\(\dfrac{\left(2\cdot3\cdot4\cdot....\cdot11\right)}{\left(1\cdot2\cdot3\cdot4\cdot...\cdot10\right)}\cdot\dfrac{\left(1\cdot2\cdot3\cdot4\cdot5\cdot...\cdot10\right)}{\left(1\cdot2\cdot3\cdot4\cdot...\cdot10\right)}\cdot x=-20\)
\(11\cdot x=-20\\ x=-\dfrac{20}{11}\)
\(2\cdot3^x-1=53\\ \Leftrightarrow2\cdot3^x=54\\ \Leftrightarrow3^x=27\\ \Leftrightarrow x=3\)
...\(\Rightarrow15-6x=3^6:3^5\)
\(\Rightarrow15-6x=3\)
\(\Rightarrow6x=12\)
\(\Rightarrow x=2\)
(15-6x).35=36
(15-6x) =36:35
(15-6x) =31
15-6x =3
6x =15-3
6x =12
x =12:6
x = 2
Vậy x=2
\(...\Rightarrow\left(2x-6\right)=4^9:4^7\)
\(\Rightarrow2x-6=4^2\)
\(\Rightarrow2x-6=16\)
\(\Rightarrow2x=22\)
\(\Rightarrow x=11\)
(2x-6). 47 = 49
2x-6 = \(\dfrac{4^9}{4^7}\)
2x = 16+6
x = \(\dfrac{22}{2}\)
x = 11
75 : (x - 18) = 52
75 : (x - 18) = 25
x - 18 = 75 : 25
x - 18 = 3
x = 3 + 18
x = 21
\(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{\left(2x-1\right)\cdot\left(2x+1\right)}=\dfrac{49}{99}\)
\(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{\left(2x-1\right)\cdot\left(2x+1\right)}=\dfrac{98}{99}\)
\(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+....+\dfrac{1}{2x-1}-\dfrac{1}{2x+1}=\dfrac{98}{99}\)
\(1-\dfrac{1}{2x+1}=\dfrac{98}{99}\)
\(\dfrac{2x+1-1}{2x+1}=\dfrac{98}{99}\)
\(\dfrac{2x}{2x+1}=\dfrac{98}{99}\)
=> 2x=98
=> x=49