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(x+1) + (x+3) + ... + (x+99) = 5100
(x+x+..+x) + (1+3+...+99) = 5100
xét dãy số 1;3;...;99 có:
(99-1) : 2 + 1 = 50 (số)
có: (x.50) + (99+1) . 50 : 2 = 5100
(x.50) + 100 .50 : 2 = 5100
(x.50) + 5000 : 2 = 5100
(x.50) + 2500 = 5100
x.50 = 5100 - 2500
x.50 = 2600
x = 2600 : 50
x = 52
vậy......
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x+1\right)+\left(x+3\right)+...+\left(x+99\right)=5100\)
\(\Leftrightarrow50x+\frac{\left[\left(99-1\right):2+1\right]\left(99+1\right)}{2}=5100\)
\(\Leftrightarrow50x+2500=5100\)
\(\Leftrightarrow50x=2600\)
\(\Leftrightarrow x=52\)
Vậy ...
\(\left(x+1\right)+\left(x+3\right)+...+\left(x+99\right)=5100.\)
\(\Leftrightarrow x+1+x+3+...+x+99=5100.\)'
\(\Leftrightarrow\left(x+x+...+x\right)+\left(1+3+...+99\right)=5100\)
50 số hạng x 50 số hạng
\(\Leftrightarrow50x+\frac{\left(1+99\right)\cdot50}{2}=5100\)
\(\Leftrightarrow50x=5100-2500=2600\)
\(\Leftrightarrow x=52\)
Vậy x = 52
Học tốt nhé ^3^
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a)(x+1)+(x+2)+...+(x+30)=795
(x+x+...+x)+(1+2+3+...+30)=795
30x+465=795
30x=795-465
30x=330
x=330/30
x=11
b)(x+1)+(x+3)+...+(x+99)=5100
(x+x+...+x)+(1+3+...+99)=5100
50x+2500=5100
50x=5100-2500
50x=2600
x=2600/50
x=52
![](https://rs.olm.vn/images/avt/0.png?1311)
4) (x+1)+(x+2)+(x+3)+....+(x+30)=795
<=> (x+x+x+...+x)+(1+2+3+....+30)=795
<=> 30x+\(\frac{\left(30+1\right)\cdot30}{2}=795\)
<=> 30x+465=795
<=> 30x=330
<=> x=11
Vậy x=11
5) (x+1)+(x+3)+(x+5)+...+(x+99)=5100
<=> (x+x+x+...+x)+\(\frac{\left(99+1\right)\cdot50}{2}=5100\)
<=> 50x+2500=5100
<=> 50x=2600
<=> x=52
Vậy x=52
\(a.15-\left|x-3\right|=7\)
\(\Rightarrow\left|x-3\right|=8\)
\(\Rightarrow\orbr{\begin{cases}x-3=8\\x-3=-8\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=11\\x=-5\end{cases}}\)
\(b.x-\left(17-x\right)=x-10\)
\(\Rightarrow17-x=x-x+10\)
\(\Rightarrow17-x=10\)
\(\Rightarrow x=7\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2:
Ta có: (x-3)(x+4)>0
=>x>3 hoặc x<-4
Bài 3:
a: \(5S=5-5^2+...+5^{99}-5^{100}\)
\(\Leftrightarrow6S=1-5^{100}\)
hay \(S=\dfrac{1-5^{100}}{6}\)