Đáp án:

 l) \(\left[ \begin{array}{l}
x = 3\\
x = \dfrac{5}{2}
\end{array} \right.\)

Giải thích các bước giải:

\(\begin{array}{l}
e){x^2} - 8x - 6x + 48 = 0\\
 \to x\left( {x - 8} \right) - 6\left( {x - 8} \right) = 0\\
 \to \left( {x - 8} \right)\left( {x - 6} \right) = 0\\
 \to \left[ \begin{array}{l}
x = 8\\
x = 6
\end{array} \right.\\
f)5{x^2} - 25x - 4x + 20 = 0\\
 \to 5\left( {x - 5} \right) - 4\left( {x - 5} \right) = 0\\
 \to \left( {x - 5} \right)\left( {x - 4} \right) = 0\\
 \to \left[ \begin{array}{l}
x = 5\\
x = 4
\end{array} \right.\\
g)\Delta  = 49 + 20 = 69\\
 \to \left[ \begin{array}{l}
x = \dfrac{{7 + \sqrt {69} }}{2}\\
x = \dfrac{{7 - \sqrt {69} }}{2}
\end{array} \right.\\
h)3{x^2} + 6x + x + 2 = 0\\
 \to 3x\left( {x + 2} \right) + \left( {x + 2} \right) = 0\\
 \to \left( {x + 2} \right)\left( {3x + 1} \right) = 0\\
 \to \left[ \begin{array}{l}
x =  - 2\\
x =  - \dfrac{1}{3}
\end{array} \right.\\
i)\Delta ' = 9 - 39 =  - 30 < 0\\
 \to x \in \emptyset \\
j)3{x^2} + 6x + 2x + 4 = 0\\
 \to 3x\left( {x + 2} \right) + 2\left( {x + 2} \right) = 0\\
 \to \left( {x + 2} \right)\left( {3x + 2} \right) = 0\\
 \to \left[ \begin{array}{l}
x =  - 2\\
x =  - \dfrac{2}{3}
\end{array} \right.\\
k) - 3{x^2} + 3x - x + 1 = 0\\
 \to  - 3x\left( {x - 1} \right) - \left( {x - 1} \right) = 0\\
 \to \left( {x - 1} \right)\left( { - 3x - 1} \right) = 0\\
 \to \left[ \begin{array}{l}
x = 1\\
x =  - \dfrac{1}{3}
\end{array} \right.\\
l)2{x^2} - 6x - 5x + 15 = 0\\
 \to 2x\left( {x - 3} \right) - 5\left( {x - 3} \right) = 0\\
 \to \left( {x - 3} \right)\left( {2x - 5} \right) = 0\\
 \to \left[ \begin{array}{l}
x = 3\\
x = \dfrac{5}{2}
\end{array} \right.
\end{array}\)

$e)\,{{x}^{2}}-14x+48=0$
$\Delta ={{\left( -14 \right)}^{2}}\,-\,4.1.48=4\to \sqrt{\Delta }=2$

$\to\left[\begin{array}{1}{{x}_{1}}=\dfrac{-b+\sqrt{\Delta }}{2a}=\dfrac{-\left( -14 \right)+2}{2.1}=8\\\\{{x}_{2}}=\dfrac{-b-\sqrt{\Delta }}{2a}=\dfrac{-\left( -14 \right)-2}{2.1}=6\end{array}\right.$

$f)\,5{{x}^{2}}-29x+20=0$
$\Delta ={{\left( -29 \right)}^{2}}-4.5.20=441\to \sqrt{\Delta }=21$

$\to\left[\begin{array}{1}{{x}_{1}}=\dfrac{-b+\sqrt{\Delta }}{2a}=\dfrac{-\left( -29 \right)+21}{2.5}=5\\\\{{x}_{2}}=\dfrac{-b-\sqrt{\Delta }}{2a}=\dfrac{-\left( -29 \right)-21}{2.5}=\dfrac{4}{5}\end{array}\right.$

$g)\,{{x}^{2}}-7x-5=0$
$\Delta ={{\left( -7 \right)}^{2}}-4.1.\left( -5 \right)=69\to \sqrt{\Delta }=\sqrt{69}$

$\to\left[\begin{array}{1}{{x}_{1}}=\dfrac{-b+\sqrt{\Delta }}{2a}=\dfrac{-\left( -7 \right)+\sqrt{69}}{2.1}=\dfrac{7+\sqrt{69}}{2}\\\\{{x}_{2}}=\dfrac{-b-\sqrt{\Delta }}{2a}=\dfrac{-\left( -7 \right)-\sqrt{69}}{2.1}=\dfrac{7-\sqrt{69}}{2}\end{array}\right.$

$h)\,3{{x}^{2}}+7x+2=0$
$\Delta ={{7}^{2}}-4.3.2=25\to \sqrt{\Delta }=5$

$\to\left[\begin{array}{1}{{x}_{1}}=\dfrac{-b+\sqrt{\Delta }}{2a}=\dfrac{-7+5}{2.3}=-\dfrac{1}{3}\\\\{{x}_{2}}=\dfrac{-b-\sqrt{\Delta }}{2a}=\dfrac{-7-5}{2.3}=-2\end{array}\right.$

$i)\,{{x}^{2}}+6x+39=0$
$\Delta ={{6}^{2}}-4.39=-120<0$
$\to $ phương trình vô nghiệm

$j)\,3{{x}^{2}}+8x+4=0$
$\Delta ={{8}^{2}}-4.3.4=16\to \sqrt{\Delta }=4$

$\to\left[\begin{array}{1}{{x}_{1}}=\dfrac{-b+\sqrt{\Delta }}{2a}=\dfrac{-8+4}{2.3}=-\dfrac{2}{3}\\\\{{x}_{2}}=\dfrac{-b-\sqrt{\Delta }}{2a}=\dfrac{-8-4}{2.3}=-2\end{array}\right.$

$k)\,-3{{x}^{2}}+2x+1=0$
$\Delta ={{2}^{2}}-4.\left( -3 \right).1=16\to \sqrt{\Delta }=4$

$\to\left[\begin{array}{1}{{x}_{1}}=\dfrac{-b+\sqrt{\Delta }}{2a}=\dfrac{-2+4}{2.\left( -3 \right)}=-\dfrac{1}{3}\\\\{{x}_{2}}=\dfrac{-b-\sqrt{\Delta }}{2a}=\dfrac{-2-4}{2.\left( -3 \right)}=1\end{array}\right.$

$l)\,2{{x}^{2}}-11x+15=0$
$\Delta ={{\left( -11 \right)}^{2}}-4.2.15=1\to \sqrt{\Delta }=1$

$\left[\begin{array}{1}{{x}_{1}}=\dfrac{-b+\sqrt{\Delta }}{2a}=\dfrac{-\left( -11 \right)+1}{2.2}=3\\\\{{x}_{2}}=\dfrac{-b-\sqrt{\Delta }}{2a}=\dfrac{-\left( -11 \right)-1}{2.2}=\dfrac{5}{2}\end{array}\right.$