a. (x+1) + (x+3) + (x+5) + ......+(x+99)=0

    x.50+(1+3+5+.......+99)=0

     x.50 + 2500                =0

     x.50                           =0+2500

      x.50                           = 2500

      x                                 =2500:50

     x                                  =50

(2x - 1) (y - 4) = 13

\(\orbr{\begin{cases}2x-1=3\\y-4=3\end{cases}}\)

\(\orbr{\begin{cases}2x=4\\y=7\end{cases}}\)

\(\orbr{\begin{cases}x=2\\y=7\end{cases}}\)

Vậy x = 2 và y = 7

k minh voi minh t lai

\(\Leftrightarrow\left[{}\begin{matrix}5x-4=x+2\\4-5x=x+2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=6\\6x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)

|5x-4|=|x+2|

TH1: 5x-4=x+2⇒4x=6⇒x=3/2

TH2: 5x-4=-x-2⇒ 6x=2⇒x=1/3

Vậy ....

\(\left|5x-4\right|=\left|x+2\right|\)

\(\Rightarrow\left[{}\begin{matrix}5x-4=x+2\\5x-4=-x-2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}5x-x=4+2\\5x+x=4-2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}4x=6\\6x=2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy \(x=\dfrac{3}{2}\) hoặc \(x=\dfrac{1}{3}\)

a]={23;24;25;26....;35}

b]={4;8;14;...}

c]{4}

Tim x thuoc N=thi ko biet nha!