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\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{1}{x\left(x+1\right):2}=\frac{2001}{2003}\)
\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{x\left(x+1\right)}=\frac{2001}{2003}\)
\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{x\left(x+1\right)}\right)=\frac{2001}{2003}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{2001}{2003}:2=\frac{2001}{4006}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(\frac{1}{x+1}=\frac{1}{2}-\frac{2001}{4006}=\frac{1}{2003}\)
=> x+1 = 2003
=> x = 2003 - 1
=> x = 2002
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x+(x+1)+(x+2)+(x+3)+.......+(x+30)=1240
\(\Leftrightarrow\left(x+x+x.+...x\right)+\left(1+2+3...+30\right)=1240\)
\(\Rightarrow30x+465=1240\)
\(\Rightarrow30x=1240-465=775\)
\(\Rightarrow30x=775\)
\(V\text{ậy}x=\frac{155}{6}\)
1+2+3+.....+x=210
\(\left(1+x\right).x=210\)
\(\Rightarrow x=14\)
x+(x+1)+(x+2)+...+(x+30)=1240
=>x+x+1+x+2+...+x+30=1240
=>(x+x+x+...+x)+(1+2+...+30)=1240
=>31x+[(30-1):1+1] . (30+1) :2=1240
=>31x+30.31:2=1240
=>31x+15.31=1240
=>31(x+15)=1240
=>x+15=1240:31=40
=>x=40-15=25
1+2+3+...+x=210
=>[(x-1):1+1]. (x+1) : 2= 210
=>x.(x+1):2=210
=>x(x+1)=210.2=420
=>x(x+1)=20.21
=>x=20
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Nhiều số lắm bạn ơi! Phải có thêm điều kiện gì chứ nhỉ?
3x+2 + 3x+1 + 3x < 1053
=> 3x+2 + 3x+1 + 3x < 36 + 35 + 34
=> x < 4
=> x thuộc {0; 1; 2; 3}
Vậy...
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a) 1^3+2^3+3^3+...+10^3
= (1+2+3+...+10)^2
= 55^2
Mặt khác: 1^3+2^3+3^3+...+10^3=(x+1)^2
hay 55^2=(x+1)^2
=> x+1=55
=> x=54
Vậy x=54
b) 1+3+5+...+99
=(1+99).50/2
=2500
=50^2
Mặt khác: 1+3+5+...+99=(x-2)^2
hay 50^2=(x-2)^2
=> x-2=50
=>x=52
Vậy x=52
k giúp mình nha!!!
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a: =>n*5^3=5^7
=>n=5^4=625
c: \(\Leftrightarrow2\cdot3^n=3^4+2^5-5=81+32-5=108\)
=>3^n=54
=>\(n\in\varnothing\)
d: =>5^n=25
=>n=2
f: =>3n+1=4
=>3n=3
=>n=1
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a: \(\Leftrightarrow x-3\inƯ\left(21\right)\)
\(\Leftrightarrow x-3\in\left\{-3;-1;1;3;7;21\right\}\)
hay \(x\in\left\{0;2;4;6;10;24\right\}\)
b: \(\Leftrightarrow x-1\in\left\{1;-1;17\right\}\)
hay \(x\in\left\{2;0;18\right\}\)
c: \(\Leftrightarrow2x-1+4⋮2x-1\)
\(\Leftrightarrow2x-1\in\left\{1;-1\right\}\)
hay \(x\in\left\{1;0\right\}\)
d: \(\Leftrightarrow x^2+x+3⋮x+1\)
\(\Leftrightarrow x+1\inƯ\left(3\right)\)
\(\Leftrightarrow x+1\in\left\{1;3\right\}\)
hay \(x\in\left\{0;2\right\}\)
có công thức nè bạn
13+23+...+n3=n.(n+1).(2n+1);6