Đáp án:

 $a) |3-2x|=x-2$

$⇒ |3-2x|=$ $\left \{ {{\text{3-2x $\geq$ 0 hay x $\geq$ $\dfrac{3}{2}$}} \atop {\text{-(3-2x) <0 hay x < $\dfrac{3}{2}$}}} \right.$ 

$\text{*Với x ≥ $\dfrac{3}{2}$ :}$

$3-2x=x-2$

$⇔3-2x-x+2=0$

$⇔ -3x+5=0$

$⇔ -3x=-5$

$⇔ \dfrac{5}{3}$ $(tm)$

$\text{Với x < $\dfrac{3}{2}$:}$

$-3+2x= x-2$

$⇔ -3+2x-x+2=0$

$⇔ x-1=0$

$⇔ x=1$ $(tm)$

$\text{Vậy S= { $\dfrac{5}{3};1$}}$

$b) |4-x|=x^2+3x-1$

$⇒ |4-x|=$ $\left \{ {{\text{4-x $\geq$ 0 hay x ≤ 4}} \atop {\text{-(4-x)>0 hay x > 4}}} \right.$ 

$\text{*Với x ≤4:}$

$4-x=x^2+3x-1$

$⇔ 4-x-x^2-3x+1 =0$

$⇔ -x^2-4x+5=0$

$⇔ -(x^2+4x-5)=0$

$⇔ -(x^2+5x-x-5)=0$

$⇔ -[x.(x-1)+5.(x-1)]=0$

$⇔ -(x+5).(x-1)=0$

$⇔$\(\left[ \begin{array}{l}-(x+5)=0\\x-1=0\end{array} \right.\) $⇔$ \(\left[ \begin{array}{l}x=-5(tm)\\x=1(tm)\end{array} \right.\) 

$\text{*Với x >4:}$

$-4+x=x^2+3x-1$

$⇔ x^2+3x-1+4-x=0$

$⇔ x^2+2x+3=0$ $(L)$

$\text{Vậy S={-5;1}}$

 

Giải thích các bước giải:

`a)``|3-2x|=x-2`

ĐK: $x-2 \geqslant 0 ⇔x \geqslant 2$

Phương trình tương đương

\(\left[ \begin{array}{l}3-2x=x-2\\3-2x=-(x-2)\end{array} \right.\)

`<=>`\(\left[ \begin{array}{l}-2x-x=-2-3\\3-2x=-x+2\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}-3x=-5\\-2x+x=2-3\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}x=\dfrac{-5}{-3}\\-x=-1\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}x=\dfrac{5}{3}\\x=1\end{array} \right.\) `(ktmđk)`

Vậy phương trình vô nghiệm.

`b)``|4-x|=x^2 +3x-1`

+) Với $4-x \geqslant 0⇔x \leqslant 4$

Phương trình tương đương:

`4-x=x^2 +3x-1`

`<=>x^2 +3x-1-4+x=0`

`<=>x^2 +4x-5=0`

`<=>x^2 +5x-x-5=0`

`<=>x(x+5)-(x+5)=0`

`<=>(x+5)(x-1)=0`

`<=>`\(\left[ \begin{array}{l}x+5=0\\x-1=0\end{array} \right.\)`<=>`\(\left[ \begin{array}{l}x=-5\\x=1\end{array} \right.\) `(tmđk)`

+) Với $4-x < 0⇔x > 4$

Phương trình tương đương:

`-(4-x)=x^2 +3x-1`

`<=>x^2 +3x-1=-4+x`

`<=>x^2 +3x-1+4-x=0`

`<=>x^2 +2x+3=0`

`<=>x^2 +2x+1+2=0`

`<=>(x+1)^2 +2=0` `\text{vô lí}`

Vì: `(x+1)^2` $\geqslant$ `0 AA x`

`->(x+1)^2 +2` $\geqslant$ `2 > 0 AA x`

`->` loại.

Vậy `S={-5;1}`