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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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a) x = 0
b) x = -1
c) x = -9
d) x = 24
e) x = 2 hoặc x = -4
f) x = 5 hoặc x = -3
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a: =>x-3=9
=>x=12
b: =>10-x=-26
=>x=36
c: =>x:4-1=2
=>x:4=3
=>x=12
d: =>x^2=4
=>x=2 hoặc x=-2
e: =>(x-2)^2=100
=>x-2=10 hoặc x-2=-10
=>x=12 hoặc x=-8
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Bài 2:
\(\Leftrightarrow5a+14\in\left\{2;3;5;7;11;13;17;19;23;29;31;37\right\}\)
\(\Leftrightarrow5a\in\left\{5;15\right\}\)
hay a=3(vì a là số nguyên tố)
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\(x+\left(x+1\right)+\left(x+2\right)+...+19+20=20\)
\(\Leftrightarrow x+\left(x+1\right)+\left(x+2\right)+...+19=0\)
\(\Leftrightarrow\frac{x\left(19+x\right)}{2}=0\)
\(\Leftrightarrow x\left(x+19\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+19=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-19\end{cases}}\)
Mà \(x\ne0\)nên \(x=-19\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\Leftrightarrow x=\dfrac{16\cdot6}{4}=24\)
b: \(\Leftrightarrow x=\dfrac{-5\cdot42}{14}=-15\)
c: \(\Leftrightarrow\dfrac{x}{11}=\dfrac{-3}{11}\)
hay x=-3
\(x+\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+14\right)=14\)
\(x+x+1+x+2+x+3+...+x+14=14\)
\(15x+\left(1+2+3+...+14\right)=14\)
\(15x+\frac{\left(14+1\right).14}{2}=14\)
\(15x+105=14\)
\(15x=-91\)
\(x=-\frac{91}{15}=-6,06\)
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