a)

\(x+\left(x+2\right)+\left(x+4\right)+...+\left(x+98\right)=0\)

\(x+x+2+x+4+...+x+98=0\)

\(50x+\left(98+2\right).\left[\left(98-2\right):2+1\right]:2=0\)

\(50x+100.49:2=0\)

\(50x+49.50=0\)

\(50x=0-49.50\)

\(50x=-2450\)

\(x=-2450:50\)

\(x=-49\)

b)

\(\left(x-5\right)+\left(x-4\right)+\left(x-3\right)+...+\left(x+11\right)+\left(x+12\right)=99\)

\(x+x+x+...+x-5-4-3-...+11+12=99\)

\(18x+6+7\text{+ 8 + 9 + 10 + 11 + 12 = 99}\)

\(18x+63=99\)

\(18x=99-63\)

\(18x=36\)

\(x=36:18\)

\(x=2\)

giúp mình với, mình đang vội!

a) x + (x + 2) + (x + 4) + ... + (x + 98) = 0

x + x + 2 + x + 4 + ... + x + 98 = 0

50x + (98 + 2).[(98 - 2) : 2 + 1]:2 = 0

50x + 100 .49 : 2 = 0

50x + 49.50 = 0

50x = 0 - 49.50

50x = -2450

x = -2450 : 50

x = -49

b) (x - 5) + (x - 4) + (x - 3) + ... + (x + 11) + (x + 12) = 99

x + x + x + ... + x - 5 - 4 - 3 - ... + 11 + 12 = 99

18x + 6 + 7 + 8 + 9 + 10 + 11 + 12 = 99

18x + 63 = 99

18x = 99 - 63

18x = 36

x = 36 : 18

x = 2

a: \(\Leftrightarrow\dfrac{x}{-4}=\dfrac{21}{y}=\dfrac{z}{-80}=\dfrac{3}{4}\)

=>x=-3; y=28; z=-60

b: 5/12=x/-72

=>x=-72*5/12=-6*5=-30

c: =>x+3=-5

=>x=-8

a, | x - 2012| =1

+) x-2012=1                       +) x-2012=-1

=> x=2013                             => x=2011

b, <=> x khác x với mọi x thuộc Z

a)\(\left|x-2012\right|-1=0=>\left|x-2012\right|=1\)

\(=>\orbr{\begin{cases}x-2012=1\\x-2012=-1\end{cases}=>\orbr{\begin{cases}x=2013\\x=2011\end{cases}}}\)

b)\(\left|x\right|>x=>x< 0\)

a, dễ dồi khỏi cần làm

b, + Với x \(\in\)Z và x \(\ge\)0 ta có : |x| = x ( loại)

    + Với x \(\in\)Z  và x < 0 ta có |x| = -x

Vì x < 0 => -x > 0 => -x > x => |x| > x ( chọn)

KL:  Vậy x \(\in\)Z và x < 0

b: x=-3/4

\(\left(-2012\right).\left(x+3\right)=0\)

\(\Leftrightarrow x+3=0\)

\(\Leftrightarrow x=-3\)

(-2012).(x+3)=0

=> x+3=0 => x=-3

\(a,\left(x+3\right)\left(5-x\right)=0\\ \Rightarrow\left\{{}\begin{matrix}x+3=0\\5-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-3\\x=5\end{matrix}\right.\)

\(c,x+17⋮x+3\\ x+3+14⋮x+3\\ 14⋮x+3\\ x+3\inƯ\left(14\right)=\left\{\pm14;\pm7\pm2;\pm1\right\}\)

Từ đó bạn tìm những giá trị của x nha!

a) \(x\left(x-6\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)

b) \(\left(-7-x\right)\left(-x+5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}-7-x=0\\-x+5=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-7\\x=-5\end{matrix}\right.\)

c) \(\left(x+3\right)\left(x-7\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x-7=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=7\end{matrix}\right.\)

d) \(\left(x-3\right)\left(x^2+12\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x^2+12=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\x^2=-12\text{(vô lý)}\end{matrix}\right.\)

\(\Rightarrow x=3\)

e) \(\left(x+1\right)\left(2-x\right)\ge0\)

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x+1\ge0\\2-x\ge0\end{matrix}\right.\\\left[{}\begin{matrix}x+1\le0\\2-x\le0\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\ge-1\\x\le2\end{matrix}\right.\\\left[{}\begin{matrix}x\le-1\\x\ge2\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}-1\le x\le2\\x\in\varnothing\end{matrix}\right.\)

\(\Rightarrow-1\le x\le2\)

f) \(\left(x-3\right)\left(x-5\right)\le0\)

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x-3\le0\\x-5\ge0\end{matrix}\right.\\\left[{}\begin{matrix}x-3\ge0\\x-5\le0\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\le3\\x\ge5\end{matrix}\right.\\\left[{}\begin{matrix}x\ge3\\x\le5\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow3\le x\le5\)

a) =>\(\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)

b => \(\left[{}\begin{matrix}-7-x=0\\-x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-7\\x=5\end{matrix}\right.\)

d) => \(\left[{}\begin{matrix}x-3=0\\x^2+12=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x^2=-12\end{matrix}\right.\)(vô lí) => x=3

c) => \(\left[{}\begin{matrix}x+3=0\\x-7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-3\\x=7\end{matrix}\right.\)