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1, -15 + \(x\) = -14 - (-57)
-15 + \(x\) = -14 + 57
\(x\) = -14 + 57 + 15
\(x\) = 43 + 15
\(x\) = 58
2, (-14) + \(x\) - 7 = -10
-14 + \(x\) - 7 = -10
- 21 + \(x\) = - 10
\(x\) = -10 + 21
\(x\) = 11
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\(x+\left(x+1\right)+\left(x+2\right)+....+13+14=14\Leftrightarrow x+\left(x+1\right)+....+13\Leftrightarrow x=-13\)
\(25+24+23+....+x+\left(x-1\right)+\left(x-2\right)=25\Leftrightarrow24+....+\left(x-2\right)=0\Leftrightarrow x-2=-24\)
\(\Leftrightarrow x=-22\)
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\(x+\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+14\right)=14\)
\(x+x+1+x+2+x+3+...+x+14=14\)
\(15x+\left(1+2+3+...+14\right)=14\)
\(15x+\frac{\left(14+1\right).14}{2}=14\)
\(15x+105=14\)
\(15x=-91\)
\(x=-\frac{91}{15}=-6,06\)
Ko chắc lắm
x+(x+1)+(x+2)+(x+3)+...+(x+13)+(x+14)=14
x+x+1+x+2+x+3+...+x+13+x+14=14
(x+x+x+x+...+x+x)+(1+2+3+...+14)=14
15x + 14 = 14
15x = 14 - 14
15x = 0
x = 0:15
x = 0
Vậy x=0
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a: =>14x+2*(14*13/2)=14
=>14x+14*13=14
=>14x=-12*14
=>x=-12
b: =>3x+(0-1-2-3+1+2+...+25)=25
=>3x+4+...+25=25
=>3x+4+...+24=0
=>3x+(24-4+1)*28/2=0
=>3x+21*28/2=0
=>3x=-21*14
=>x=-7*14=-98
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Ta có : \(\frac{1}{2}+\frac{1}{14}+\frac{1}{35}+...+\frac{1}{x\left(x+3\right):2}=\frac{9}{14}\)
\(\Rightarrow2\left(\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+...+\frac{1}{x\left(x+3\right)}\right)=\frac{9}{14}\)
\(\Rightarrow2.\frac{1}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{x\left(x+3\right)}\right)=\frac{9}{14}\)
\(\Rightarrow\frac{2}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{9}{14}\)
\(\Rightarrow1-\frac{1}{x+3}=\frac{27}{28}\)
\(\Rightarrow\frac{1}{x+3}=\frac{1}{28}\)
=> x + 3 = 28
=> x = 25
![](https://rs.olm.vn/images/avt/0.png?1311)
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![](https://rs.olm.vn/images/avt/0.png?1311)
x+(x+1) +(x+2) + ... + (x+14 )=14
(x+x+x+...+x) + (1+2+3+4+ ...+14)=14
15x + [(14-1):1+1].(1+14) :2=14
15x + 105=14
15x=-91
suy ra x=-91/15
x + 14 chia hết x + 1
=> x + 1 + 13 chia hết cho x + 1
=> 13 chia hết cho x + 1
=> x + 1 thuộc Ư(13)={±1;±13}
=> x thuộc { 0;-2;12;-14 }
ko cs dau = thi lam kieu j