a) Ix + 1I = x - 2
<=> x + 1 = x - 2         hay         x + 1 = 2 - x
<=> x - x = -2 - 1           I <=>    x + x = 2 - 1
<=>   0x  = -3 (vô lí)       I <=>      2x   = 1
                                    I <=>       x    = 1/2

b) Ix - 1I = I2xI (*)

TH1: x < 0
(*) <=> 1 - x = -2x
    <=>  -x + 2x = -1
    <=>      x      = -1
TH2: 0 <= x < 1
(*) <=> 1 - x = 2x
    <=> -x - 2x = -1
    <=>   - 3x   = -1
   <=>        x   = 1/3
TH3: x >= 1
(*) <=> x - 1 = 2x
    <=> x - 2x = 1
    <=>   -x    = 1
    <=>    x    = -1

c) Ix - 3I + Ix - 2I = 4 (**)

TH1: x < 2
(**) <=> 3 - x + 2 - x = 4
      <=>       -2x       = 4 - 3 - 2
      <=>       -2x       = -1
      <=>          x       = 1/2
TH2: 2 <= x < 3
(**) <=> 3 - x + x - 2 = 4
      <=>       0x        = 4 + 2 + 3
      <=>       0x        = 9 (vô lí)
TH3: x >= 3
(**) <=> x - 3 + x - 2 = 4
      <=>        2x       = 4 + 2 + 3
      <=>        2x       = 9
      <=>          x       = 9/2

A, bạn ơi mình biết làm hết r

B,cài này dễ lằm bạn nghĩ đi okeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee

ko có đáp án

a: \(\Leftrightarrow\left[{}\begin{matrix}2x-5=3-8x\\2x-5=8x-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}10x=8\\-6x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=-\dfrac{1}{3}\end{matrix}\right.\)

a: \(\Leftrightarrow\left\{{}\begin{matrix}3x-2>-4\\3x-2< 4\end{matrix}\right.\Leftrightarrow-\dfrac{2}{3}< x< 2\)

c: \(\Leftrightarrow\left[{}\begin{matrix}3x-1>5\\3x-1< -5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>2\\x< -\dfrac{4}{3}\end{matrix}\right.\)

d: \(\Leftrightarrow\left[{}\begin{matrix}3x+1>x-2\\3x+1< -x+2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x>-3\\4x< 1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>-\dfrac{3}{2}\\x< \dfrac{1}{4}\end{matrix}\right.\)