Đáp án:

 $1, 2x(x-7) +5x-35=0$

$⇔2x(x-7) +5(x-7)=0$

$⇔(x-7)(2x+5)=0$

$⇔$\(\left[ \begin{array}{l}x=7\\x=-\dfrac{5}{2}\end{array} \right.\) 

$\text{ Vậy x ∈ {$7 ; -\dfrac{5}{2}$}}$

$2, x(x-3) -7x+21 =0$

$⇔x(x-3)-7(x-3)=0$

$⇔(x-3)(x-7)=0$

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

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

$\text{Vậy x ∈ { 3 ; 7 }}$

$3, x^3 -2x^2 +x-2=0$

$⇔x^2(x-2)+(x-2)=0$

$⇔(x-2)(x^2+1)=0$

Vì $x^2 +1 > 0 $ nên ta loại

$⇔x-2=0⇔x=2$

Vậy $x=2$

$4, x^3 -5x^2-x+5=0$

$⇔x^2(x-5)-(x-6)=0$

$⇔(x-5)(x^2-1)=0$

$⇔(x-5)(x-1)(x+1)=0$

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

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

$\text{ Vậy x ∈ {5 ; 1 ; -1}}$

$5, x^2 -6x+8=0$

$⇔x^2 -2x -4x +8=0$

$⇔x(x-2)-4(x-2)=0$

$⇔(x-2)(x-4)=0$

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

$⇔$\(\left[ \begin{array}{l}x=2\\x=4\end{array} \right.\) 

$\text{Vậy x ∈ {2 ; 4} }$

$6, 9x^2 +6x-3=0$

$⇔9x^2+9x-3x-3=0$

$⇔9x(x+1)-3(x+1)=0$

$⇔(x+1)(9x-3)=0$

$⇔$\(\left[ \begin{array}{l}x+1=0\\9x-3=0\end{array} \right.\) 

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

$\text{Vậy x ∈ {$-1 ;\dfrac{1}{3}$ }}$

$7, x^3 +x^2+x+1=0$

$⇔x^2(x+1) +(x+1)=0$

$⇔(x+1)(x^2+1)=0$

Vì $x^2+1 >0$ nên ta loại

$⇔x+1=0⇔x=-1$

Vậy $x=-1$

$8, x^3 -x^2-x+1=0$

$⇔x^2(x-1)-(x-1)=0$

$⇔(x-1)(x^2-1)=0$

$⇔x-1=0⇔x=1$

$⇔x^2 =1⇔x=±1$

$\text{Vậy x ∈ {1 ; -1}}$

$9, 4x^2 -49=0$

$⇔(2x-7)(2x+7)=0$

$⇔$\(\left[ \begin{array}{l}2x-7=0\\2x+7=0\end{array} \right.\) 

$⇔$\(\left[ \begin{array}{l}x=\dfrac{7}{2}\\x=-\dfrac{7}{2}\end{array} \right.\) 

$\text{Vậy x ∈ { $\dfrac{7}{2} ; -\dfrac{7}{2}$}}$

$10, x^2+36=0$

Vì $x^2 ≥ 0$

Nên $x^2+36 > 0$

Vậy x  vô nghiệm

$11, x^2-25=0$

$⇔x^2=25$

$⇔x=±5$

Vậy $x=±5$

$12, x^2-4x+4=0$

$⇔(x-2)^2=0$

$⇔x-2=0$

$⇔x=2$

Vậy $x=2$