Đáp án+ Giải thích các bước giải:

a) $x^2-25-(x+5)=0$
$⇔ (x-5)(x+5)-(x+5)=0$
$⇔ (x+5)(x-5-1)=0$
$⇔ (x+5)(x-6)=0$
⇔\(\left[ \begin{array}{l}x+5=0\\x-6=0\end{array} \right.\) <=>\(\left[ \begin{array}{l}x=-5\\x=6\end{array} \right.\) 

⇒ $\text{Tập nghiệm Pt: S = { -5;6}}$

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

⇒ $\text{ Tập nghiệm Pt: S= {2; 1/3}}$

c) $x(x-4)-2x+8=0$
$⇔ x(x-4)-2(x-4)=0$
$⇔ (x-4)(x-2)=0$
⇔\(\left[ \begin{array}{l}x-4=0\\x-2=0\end{array} \right.\) <=>\(\left[ \begin{array}{l}x=4\\x=2\end{array} \right.\) 

⇒ $\text{ Tập nghiệm PT: S ={ 4;2}}$

d) $3x(x+5)-3x-15=0$
$⇔ 3x(x+5)-3(x+5)=0$
$⇔ (x+5)(3x-3)=0$
$⇔ 3(x+5)(x-1)=0$
⇔\(\left[ \begin{array}{l}x+5=0\\x-1=0\end{array} \right.\) <=>\(\left[ \begin{array}{l}x=-5\\x=1\end{array} \right.\) 

⇒ $\text{ Tập nghiệm Pt: S= { -5;1}}$

`a)x^2-25-(x+5)=0`

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

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

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

`=>(x+5)(x-6)=0`

`=>`$\left[\begin{matrix}x+5=0\\ x-6=0\end{matrix}\right.$

`=>`$\left[\begin{matrix}x=-5\\x=6\end{matrix}\right.$

`b)3x(x-2)-x+2=0`

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

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

`=>`$\left[\begin{matrix} x-2=0\\ 3x-1=0\end{matrix}\right.$

`=>`$\left[\begin{matrix} x=2\\ 3x=1\end{matrix}\right.$

`=>`$\left[\begin{matrix} x=2\\ x=\dfrac{1}{3}\end{matrix}\right.$

`c)x(x-4)-2x+8=0`

`=>x(x-4)-(2x-8)=0`

`=>x(x-4)-2(x-4)=0`

`=>(x-4)(x-2)=0`

`=>`$\left[\begin{matrix}x-4=0\\ x-2=0\end{matrix}\right.$

`=>`$\left[\begin{matrix} x=4\\ x=2\end{matrix}\right.$

`d)3x(x+5)-3x-15=0`

`=>3x(x+5)-(3x+15)=0`

`=>3x(x+5)-3(x+5)=0`

`=>(x+5)(3x-3)=0`

`=>`$\left[\begin{matrix} x+5=0\\ 3x-3=0\end{matrix}\right.$

`=>`$\left[\begin{matrix} x=-5\\ 3x=3\end{matrix}\right.$

`=>`$\left[\begin{matrix} x=-5\\ x=\frac{3}{3}\end{matrix}\right.$

`=>`$\left[\begin{matrix} x=-5\\ x=1\end{matrix}\right.$