Question

|x-2|+x=3

Answer 1

a, \(\left|x-2\right|+x=3\)

+) Xét \(x\ge2\) có:
\(x-2+x=3\Leftrightarrow2x=5\Leftrightarrow x=2,5\) ( t/m )

+) Xét \(x< 2\) có:
\(2-x+x=3\Leftrightarrow2=3\) ( vô lí )

Vậy x = 2,5

Answer 2

\(\left|x-2\right|+x=3\)

\(\left|x+2\right|=3-x\)

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

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

\(\Rightarrow\left\{{}\begin{matrix}2x=5\\2x=-1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{-1}{2}\end{matrix}\right.\)

Vậy \(x\in\left\{\dfrac{5}{2};\dfrac{-1}{2}\right\}\)

Answer 3

\(\left|x-2\right|+3=3\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2+x=-3\left(đk:x-2\ge0\right)\\-\left(x-2\right)+x=3\left(đk:x-2< 0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\left(đk:x\ge2\right)\\x\in\varnothing\left(đk:x< 2\right)\end{matrix}\right.\)

\(\Leftrightarrow x=\dfrac{5}{2}\)

Vậy \(x=\dfrac{5}{2}\)

Answer 4

+) Nếu : \(x< 2\Rightarrow x-2< 2-2\Rightarrow x-2< 0\)

\(\Rightarrow\left|x-2\right|=-\left(x-2\right)=-x+2\)

\(\Rightarrow-x+2+x=3\Rightarrow-x+x=3-2\Rightarrow0=1\) (loại)

+) Nếu : \(x\ge2\Rightarrow x-2\ge2-2\Rightarrow x-2\ge0\)

\(\Rightarrow\left|x-2\right|=x-2\)

\(\Rightarrow x-2+x=3\Rightarrow x+x-2=3\)

\(\Rightarrow2x-2=3\Rightarrow2x=5\Rightarrow x=\dfrac{5}{2}\)

Vậy x = \(\dfrac{5}{2}\)

Answer 5

a,

\(\left|x-2\right|+x=3\)

- Xét \(x\ge2\) có :

\(x-2+x=3\)

\(\Leftrightarrow2x=5\Leftrightarrow x=2,5\) ( t/m )

- Xét \(x\le2\) có :

\(2-x+x=3\)

\(\Leftrightarrow2=3\) ( vô lý - loại )

Vậy x chỉ có thể là 2,5 .