Đáp án:

 $4) x^2-4=(x-2).(3-2x)$

$⇔ (x-2).(x+2)-(x-2).(3-2x)=0$

$⇔ (x-2).(x+2-3+2x)=0$

$⇔ (x-2).(3x-1)=0$

$⇔$ \(\left[ \begin{array}{l}x-2=0\\3x-1=0\end{array} \right.\) $⇔$ \(\left[ \begin{array}{l}x=2\\3x=1\end{array} \right.\) $⇔$ \(\left[ \begin{array}{l}x=2\\x=\dfrac{1}{3}\end{array} \right.\) 

$\text{Vậy S={2 ; $\dfrac{1}{3}$ }}$

$5) (2x+5)^2=(x+2)^2$

 $⇔ (2x+5)^2-(x+2)^2=0$

$⇔ (2x+5-x-2).(2x+5+x+2)=0$

$⇔ (x+3).(3x+7)=0$

$⇔$ \(\left[ \begin{array}{l}x+3=0\\3x+7=0\end{array} \right.\) $⇔$ \(\left[ \begin{array}{l}x=-3\\3x=-7\end{array} \right.\) $⇔$ \(\left[ \begin{array}{l}x=-3\\x=-\dfrac{7}{3}\end{array} \right.\) 

$\text{Vậy S={-3 ; $-\dfrac{7}{3}$}}$

$6) x^2-5x+6=0$

$⇔ x^2-2x-3x+6=0$

$⇔ x.(x-2)-3.(x-2)=0$

$⇔ (x-3).(x-2)=0$

$⇔$ \(\left[ \begin{array}{l}x-3=0\\x-2=0\end{array} \right.\) $⇔$ \(\left[ \begin{array}{l}x=3\\x=2\end{array} \right.\) 

$\text{Vậy S={3;2}}$

`4)`

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

`<=>(x-2)(x+2)-(x-2)(3-2x)=0`

`<=>(x-2)[(x+2)-(3-2x)]=0`

`<=>(x-2)(x+2-3+2x)=0`

`<=>(x-2)(3x-1)=0`

`<=>`\(\left[ \begin{array}{l}x-2=0\\3x-1=0\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}x=2\\x=\dfrac{1}{3}\end{array} \right.\) 

Vậy phương trình có tập nghiệm là `S={2; 1/3}`

$\\$

`5)`

`(2x+5)^2=(x+2)^2`

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

`<=>[(2x+5)-(x+2)][(2x+5)+(x+2)]=0`

`<=>(2x+5-x-2)(2x+5+x+2)=0`

`<=>(x+3)(3x+7)=0`

`<=>`\(\left[ \begin{array}{l}x+3=0\\3x+7=0\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}x=-3\\x=-\dfrac{7}{3}\end{array} \right.\) 

Vậy phương trình có tập nghiệm là `S={-3; -7/3}`

$\\$

`6)`

`x^2-5x+6=0`

`<=>x^2-2x-3x+6=0`

`<=>(x^2-2x)-(3x-6)=0`

`<=>x(x-2)-3(x-2)=0`

`<=>(x-2)(x-3)=0`

`<=>`\(\left[ \begin{array}{l}x-2=0\\x-3=0\end{array} \right.\) 

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

Vậy phương trình có tập nghiệm là `S={2; 3}`

$\textit{#Doraemon}$