$\text{Đáp án và giải thích các bước giải:}$

$\text{Bài 1:}$

$a)x^2+5x+6$

$=x^2+2x+3x+6$

$=x(x+2)+3(x+2)$

$=(x+2)(x+3)$

$b)3x^2-8x+4$

$=3x^2-6x-2x+4$

$=3x(x-2)-2(x-2)$

$=(x-2)(3x-2)$

$c)x^2+x-6$

$=x^2-2x+3x-6$

$=x(x-2)-3(x-2)$

$=(x-2)(x-3)$

$d)2x^2+3x-5$

$=2x^2+2x-5x-5$

$=2x(x+1)-5(x+1)$

$=(x+1)(2x-5)$

$e)x^2+5x-14$

$=x^2-2x+7x-14$

$=x(x-2)+7(x-2)$

$=(x-2)(x+7)$

$f)-x^2+4x+5$

$=-x^2-x+5x+5$

$=-x(x+1)+5(x+1)$

$=(x+1)(5-x)$

$\text{Bài 2:}$

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

$⇔x^2-2x-3x+6=0$

$⇔x(x-2)-3(x-2)=0$

$⇔(x-2)(x-3)=0$

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

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

$\text{Vậy S = {2; 3}}$

$b)3x^2+7x-10=0$

$⇔3x^2-3x+10x-10=0$

$⇔3x(x-1)+10(x-1)=0$

$⇔(x-1)(3x+10)=0$

$⇔\left[\begin{array}{l}x-1=0\\3x+10=0\end{array}\right.$

$⇔\left[\begin{array}{l}x=1\\x=-\frac{10}{3}\end{array}\right.$

$\text{Vậy S = {1; -$\frac{10}{3}$}}$

$d)x^2-6x+8=0$

$⇔x^2-2x-4x+8=0$

$⇔x(x-2)-4(x-2)=0$

$⇔(x-2)(x-4)=0$

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

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

$\text{Vậy S = {2; 4}}$

$e)x^2-7x+10=0$

$⇔x^2-2x-5x+10=0$

$⇔x(x-2)-5(x-2)=0$

$⇔(x-2)(x-5)=0$

$⇔\left[\begin{array}{l}x-2=0\\x-5=0\end{array}\right.$

$⇔\left[\begin{array}{l}x=2\\x=5\end{array}\right.$

$\text{Vậy S = {2; 5}}$

Đáp án:

$\begin{array}{l}
1)a){x^2} + 5x + 6\\
 = {x^2} + 2x + 3x + 6\\
 = \left( {x + 2} \right)\left( {x + 3} \right)\\
b)3{x^2} - 8x + 4\\
 = 3{x^2} - 2x - 6x + 4\\
 = \left( {3x - 2} \right)\left( {x - 2} \right)\\
c){x^2} + x - 6\\
 = {x^2} + 3x - 2x - 6\\
 = \left( {x + 3} \right)\left( {x - 2} \right)\\
d)2{x^2} + 3x - 5\\
 = 2{x^2} + 5x - 2x - 5\\
 = \left( {2x + 5} \right)\left( {x - 1} \right)\\
e){x^2} + 5x - 14\\
 = {x^2} + 7x - 2x - 14\\
 = \left( {x + 7} \right)\left( {x - 2} \right)\\
f) - {x^2} + 4x + 5\\
 =  - {x^2} + 5x - x + 5\\
 = \left( { - x + 5} \right)\left( {x + 1} \right)\\
B2)a){x^2} - 5x + 6 = 0\\
 \Leftrightarrow \left( {x - 3} \right)\left( {x - 2} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 3\\
x = 2
\end{array} \right.\\
Vậy\,x = 2;x = 3\\
b)3{x^2} + 7x - 10 = 0\\
 \Leftrightarrow 3{x^2} + 10x - 3x - 10 = 0\\
 \Leftrightarrow \left( {3x + 10} \right)\left( {x - 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  - \frac{{10}}{3}\\
x = 1
\end{array} \right.\\
Vậy\,x =  - \frac{{10}}{3};x = 1\\
d){x^2} - 6x + 8 = 0\\
 \Leftrightarrow \left( {x - 2} \right)\left( {x - 4} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x = 4
\end{array} \right.\\
Vậy\,x = 2;x = 4\\
e){x^2} - 7x + 10 = 0\\
 \Leftrightarrow \left( {x - 2} \right)\left( {x - 5} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x = 5
\end{array} \right.\\
Vậy\,x = 2;x = 5
\end{array}$