1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{10}\)
+......+ \(\dfrac{2}{x(x+1)}\) =1\(\dfrac{1989}{1991}\)
HeLp me
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a,4^8:X=4^6
X=4^8:4^6
X=4^2
b,12x-33=3^5
12x-33=243
12x=243+33
12x=276
12*X=276
X= 276:12
x=23
c,(5x+335):2=20^2
(5x+335):2=400
(5x+335)=400*2
(5x+335)=800
5*x+335=800
5*x=800-335
5*x=465
x=465:5
x=93
d, (x^2-10):5=3
x^2-10=3*5
x^2-10=15
x^2=15+10
x^2=25
x^2=5^2
vậy x=5
e,740:(x+10)=10^2 - 2*13
740:(x+10)=10^2-26
740:(x+10)=100-26
740:(x+10)=74
x+10=740:74
x+10=10
x=10-10
x=0
nhớ tik cho mik nhé
\(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{\left(2x-1\right)\cdot\left(2x+1\right)}=\dfrac{49}{99}\)
\(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{\left(2x-1\right)\cdot\left(2x+1\right)}=\dfrac{98}{99}\)
\(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+....+\dfrac{1}{2x-1}-\dfrac{1}{2x+1}=\dfrac{98}{99}\)
\(1-\dfrac{1}{2x+1}=\dfrac{98}{99}\)
\(\dfrac{2x+1-1}{2x+1}=\dfrac{98}{99}\)
\(\dfrac{2x}{2x+1}=\dfrac{98}{99}\)
=> 2x=98
=> x=49
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
\(\dfrac{2}{1^2}\cdot\dfrac{6}{2^2}\cdot\dfrac{12}{3^3}\cdot.......\cdot\dfrac{110}{10^2}\cdot x=-20\)
\(\dfrac{2}{1\cdot1}\cdot\dfrac{2\cdot3}{2\cdot2}\cdot\cdot\cdot\cdot\dfrac{11\cdot10}{10\cdot10}\cdot x=-20\)
\(\dfrac{\left(2\cdot3\cdot4\cdot....\cdot11\right)}{\left(1\cdot2\cdot3\cdot4\cdot...\cdot10\right)}\cdot\dfrac{\left(1\cdot2\cdot3\cdot4\cdot5\cdot...\cdot10\right)}{\left(1\cdot2\cdot3\cdot4\cdot...\cdot10\right)}\cdot x=-20\)
\(11\cdot x=-20\\ x=-\dfrac{20}{11}\)
no help