PLS HELPPP

1. |x2-1|=|x-1|

$\begin{array}{rcll} |x^2-1| &=& |x-1| \qquad & (x^2-1) = (x-1)(x+1) \\ (|(x-1)(x+1)|)^2 &=& (|x-1|)^2 \qquad & (\text{square both sides}) \\ (~(x-1)(x+1)~)^2 &=& (x-1)^2 \qquad & (~(ab)^2 = a^2b^2~) \\ (x-1)^2(x+1)^2 &=& (x-1)^2 \\ (x-1)^2(x+1)^2-(x-1)^2 &=& 0\\ (x-1)^2\left[(x+1)^2-1 \right] &=& 0\\ (x-1)^2(x^2+2x+1-1) &=& 0\\ (x-1)^2(x^2+2x) &=& 0\\ (x-1)^2\cdot x \cdot (x+2) &=& 0\\ (x-1)\cdot (x-1)\cdot x \cdot (x+2) &=& 0 \end{array}\\ \begin{array}{rrcl} \\ 1. & x-1=0 && x=1 \\ 2. & x = 0 \\ 3. & x+2 = 0 && x = -2 \end{array}$

2. 3|x-2|=2|x-3|

$\begin{array}{rcll} 3|x-2| &=& 2|x-3| \\ (3|x-2|)^2 &=& (2|x-3|)^2 \qquad & (\text{square both sides}) \\ 3^2(x-2)^2 &=& 2^2(x-3)^2 \qquad & (~(ab)^2 = a^2b^2~) \\ 9(x-2)^2 &=& 4(x-3)^2 \\ 9(x^2-4x+4) &=& 4(x^2-6x+9) \\ 9x^2-36x+36 &=& 4x^2-24x+36 \\ 9x^2-36x &=& 4x^2-24x\\ 5x^2-12x &=& 0\\ x\cdot(5x-12) &=& 0\\ \end{array}\\ \begin{array}{rrcl} \\ 1. & x = 0 \\ 2. & 5x-12=0 && x=\frac{12}{5} \end{array}$

3. |4-x|=5

$\begin{array}{rcll} |4-x| &=& 5 \\ (|4-x|)^2 &=& (5)^2 \qquad & (\text{square both sides}) \\ (4-x)^2 &=& 25 \qquad & (\pm\sqrt{}) \\ 4-x &=& \pm 5\\ x &=& 4 \pm 5 \\ \end{array}\\ \begin{array}{rrcl} \\ 1. & x = 4+5 && x = 9 \\ 2. & x = 4-5 && x = -1 \end{array}$

4. |2x-7|=|x-5|

$\begin{array}{rcll} |2x-7| &=& |x-5| \\ (|2x-7|)^2 &=& (|x-5|)^2 \qquad & (\text{square both sides}) \\ (2x-7)^2 &=& (x-5)^2 \\ 4x^2-28x+49 &=& x^2-10x+25\\ 3x^2-18x+24 &=& 0\qquad & :3\\ x^2-6x+8 &=& 0\\ (x-4)(x-2) &=& 0 \end{array}\\ \begin{array}{rrcl} \\ 1. & x-4=0 && x = 4 \\ 2. & x-2=0 && x=2 \end{array}$

5. |x2-x|=|x2-1|

$\begin{array}{rcll} |x^2-x| &=& |x^2-1| \\\\ \qquad x^2-x = x(x-1) \\ \qquad (x^2-1) = (x-1)(x+1) \\\\ |x(x-1)| &=& |(x-1)(x+1)| \\ (~|x(x-1)|~)^2 &=& (~|(x-1)(x+1)|~)^2 & (\text{square both sides}) \\ (~x(x-1)~)^2 &=& (~(x-1)(x+1)~)^2 & (~(ab)^2 = a^2b^2~) \\ x^2(x-1)^2 &=& (x-1)^2(x+1)^2\\ x^2(x-1)^2 - (x-1)^2(x+1)^2 &=& 0\\ (x-1)^2 [ x^2 - (x+1)^2 ] &=& 0\\ (x-1)^2 ( x^2 - x^2-2x-1 ) &=& 0\\ (x-1)^2 ( -2x-1 ) &=& 0\\ (x-1)\cdot (x-1)\cdot ( -2x-1 ) &=& 0\\ \end{array}\\ \begin{array}{rrcl} \\ 1. & x-1=0 && x=1 \\ 2. & -2x-1 = 0 && x = -\frac{1}{2} \end{array}$

are all the problems by itself? or am I suppose to combine them all??

These are always a little difficult to wrap one's head around.....!!!

They are all solved in pretty much the same manner  ....!!!

1.  abs (x^2 - 1) = abs (x - 1)

We have two  equations that result

x^2 - 1  =  x - 1                       and       x^2 -1  = -(x - 1)            working on both, we have

x^2  = x                                               x^2  - 1   = -x + 1

x^2 - x = 0                                          x^2 + x - 2  = 0

x(x-1)  = 0                                          (x +2) ( x-1)  = 0

So  x = 0 , 1                                      So  x = -2  and x = 1

The solutions are  x = -2, x = 0 and x = 1

Look at the graph of the solutions, here.........https://www.desmos.com/calculator/sanncuv3uw

2.  3|x-2|=2|x-3|    divide both sides by 2

(3/2)|x-2|=lx-3|      as before, we have.....

(3/2)(x - 2)  = x -3                and          (3/2)(x - 2)  =  -(x - 3)

                                                            (3/2)lx - 2 l  =  - x + 3

Multiply through by 2   in both equations.....

3(x -2) = 2x - 6                                     

3x - 6   = 4x - 6                                        3(x - 2)  = -2x + 6                               

The only solution to this is                     3x - 6    = -2x + 6

when x = 0  

                                                               5x  = 12

                                                                  x = 12/5  =  2 + 2/5  =  2.4

So the two solutions are  x = 0   and x  = 2.4

See the solution graphs, here.....   https://www.desmos.com/calculator/frf6guc36z

3. |4-x|=5       this one is easy

Either

4 - x   = 5                          or                      4 - x  = -5

-x  = -1                                                         -x   = -9

x = 1                                                              x = 9

These solutions are easily verified in the original problem

4 .   |2x-7|=|x-5|

2x - 7    =    x  - 5                    or          2x - 7   = -(x - 5)

x  =  2                                                    2x - 7  = -x  + 5

                                                              3x  = 12

                                                                x = 4

Again......these are easily verified

5.   |x^2-x|=|x^2-1|      similar to the first one, we have

x^2 - x   = x^2 - 1                  or             x^2 - x  =  - (x^2 - 1)

-x = -1                                                   x^2 - x  = -x^2 + 1

x = 1                                                   2x^2 - x - 1  = 0

                                                           (2x +1)  ( x - 1)  = 0

                                                         Set each factor to  0

                                                            x = -1/2     [ x = 1    .....a repeated solution]

Here's the graph......https://www.desmos.com/calculator/kcbcpd3khj