Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
a,\\
{x^2} - x - 3 = 0\\
 \Leftrightarrow \left( {{x^2} - x + \dfrac{1}{4}} \right) - \dfrac{1}{4} - 3 = 0\\
 \Leftrightarrow \left( {{x^2} - 2.x.\dfrac{1}{2} + \dfrac{1}{{{2^2}}}} \right) = \dfrac{{13}}{4}\\
 \Leftrightarrow {\left( {x - \dfrac{1}{2}} \right)^2} = \dfrac{{13}}{4}\\
 \Leftrightarrow \left[ \begin{array}{l}
x - \dfrac{1}{2} = \dfrac{{\sqrt {13} }}{2}\\
x - \dfrac{1}{2} = \dfrac{{ - \sqrt {13} }}{2}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{1 + \sqrt {13} }}{2}\\
x = \dfrac{{1 - \sqrt {13} }}{2}
\end{array} \right.\\
b,\\
2{x^2} + x - 7 = 0\\
 \Leftrightarrow \left( {2{x^2} + x + \dfrac{1}{8}} \right) - 7 - \dfrac{1}{8} = 0\\
 \Leftrightarrow 2.\left( {{x^2} + \dfrac{1}{2}x + \dfrac{1}{{16}}} \right) = \dfrac{{57}}{8}\\
 \Leftrightarrow 2.\left( {{x^2} + 2.x.\dfrac{1}{4} + \dfrac{1}{{{4^2}}}} \right) = \dfrac{{57}}{8}\\
 \Leftrightarrow {\left( {x + \dfrac{1}{4}} \right)^2} = \dfrac{{57}}{{16}}\\
 \Leftrightarrow \left[ \begin{array}{l}
x + \dfrac{1}{4} = \dfrac{{\sqrt {57} }}{4}\\
x + \dfrac{1}{4} =  - \dfrac{{\sqrt {57} }}{4}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{\sqrt {57}  - 1}}{4}\\
x = \dfrac{{ - \sqrt {57}  - 1}}{4}
\end{array} \right.\\
c,\\
5{x^2} + x - 9 = 0\\
 \Leftrightarrow \left( {5{x^2} + x + \dfrac{1}{{20}}} \right) - \dfrac{1}{{20}} - 9 = 0\\
 \Leftrightarrow 5.\left( {{x^2} + \dfrac{1}{5}x + \dfrac{1}{{100}}} \right) = \dfrac{{181}}{{20}}\\
 \Leftrightarrow {x^2} + 2.x.\dfrac{1}{{10}} + \dfrac{1}{{{{10}^2}}} = \dfrac{{181}}{{100}}\\
 \Leftrightarrow {\left( {x + \dfrac{1}{{10}}} \right)^2} = \dfrac{{181}}{{100}}\\
 \Leftrightarrow \left[ \begin{array}{l}
x + \dfrac{1}{{10}} = \dfrac{{\sqrt {181} }}{{10}}\\
x + \dfrac{1}{{10}} =  - \dfrac{{\sqrt {181} }}{{10}}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{\sqrt {181}  - 1}}{{10}}\\
x = \dfrac{{ - \sqrt {181}  - 1}}{{10}}
\end{array} \right.\\
d,\\
 - 3{x^2} - x + 14 = 0\\
 \Leftrightarrow 3{x^2} + x - 14 = 0\\
 \Leftrightarrow \left( {3{x^2} - 6x} \right) + \left( {7x - 14} \right) = 0\\
 \Leftrightarrow 3x.\left( {x - 2} \right) + 7.\left( {x - 2} \right) = 0\\
 \Leftrightarrow \left( {x - 2} \right)\left( {3x + 7} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x =  - \dfrac{7}{3}
\end{array} \right.\\
e,\\
 - {x^2} + x + 17 = 0\\
 \Leftrightarrow {x^2} - x - 17 = 0\\
 \Leftrightarrow \left( {{x^2} - x + \dfrac{1}{4}} \right) - 17 - \dfrac{1}{4} = 0\\
 \Leftrightarrow {\left( {x - \dfrac{1}{2}} \right)^2} = \dfrac{{69}}{4}\\
 \Leftrightarrow \left[ \begin{array}{l}
x - \dfrac{1}{2} = \dfrac{{\sqrt {69} }}{2}\\
x - \dfrac{1}{2} =  - \dfrac{{\sqrt {69} }}{2}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{\sqrt {69}  + 1}}{2}\\
x = \dfrac{{ - \sqrt {69}  + 1}}{2}
\end{array} \right.
\end{array}\)

Đáp án:

 $a) x^2 -x -3=0$

$ ⇔x^2 -2 . x . \dfrac{1}{2} + \dfrac{1}{4} -\dfrac{13}{4}=0$

$⇔ (x-\dfrac{1}{2})^2 - \dfrac{13}{4} =0$

$⇔(x-\dfrac{1}{2} -\sqrt[]{\dfrac{13}{4}}) .(x-\dfrac{1}{2}+\sqrt[]{\dfrac{13}{4}})=0$

⇔ \(\left[ \begin{array}{l}x-\dfrac{1}{2}-\sqrt[]{\dfrac{13}{4}}=0\\x-\dfrac{1}{2} +\sqrt[]{\dfrac{13}{4}}=0\end{array} \right.\) 

⇔\(\left[ \begin{array}{l}x=\dfrac{1+\sqrt[]{13}}{2}\\x=\dfrac{1-\sqrt[]{13}}{2}\end{array} \right.\) 

Vậy....

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

$⇔(√2x)^2 +2 . (√2x) . \dfrac{\sqrt[]{2}}{4} + \dfrac{1}{8} -\dfrac{57}{8}=0$

$⇔(√2x-\dfrac{\sqrt[]{2}}{4})^2 - \dfrac{57}{8} =0$

$⇔(√2x-\dfrac{\sqrt[]{2}}{4} -\sqrt[]{\dfrac{57}{8}}) .(√2x-\dfrac{\sqrt[]{2}}{4} +\sqrt[]{\dfrac{57}{8}}) =0$

⇔\(\left[ \begin{array}{l}√2x-\dfrac{\sqrt[]{2}}{4} -\sqrt[]{\dfrac{57}{8}}=0\\√2x-\dfrac{\sqrt[]{2}}{4} +\sqrt[]{\dfrac{57}{8}}=0\end{array} \right.\) 

⇔\(\left[ \begin{array}{l}x=\dfrac{1+\sqrt[]{57}}{4}\\x=\dfrac{1-\sqrt[]{57}}{4}\end{array} \right.\) 

Vậy....

$c) 5x^2 +x-9=0$

$⇔ (√5x)^2 +2 . (√5x) . \dfrac{\sqrt[]{5}}{10} + \dfrac{1}{20} -\dfrac{181}{20} =0$

$⇔(√5x -\dfrac{\sqrt[]{5}}{10})^2 - \dfrac{181}{20} =0$

$⇔(√5x-\dfrac{\sqrt[]{5}}{10} - \sqrt[]{\dfrac{181}{20}}) . (√5x -\dfrac{\sqrt[]{5}}{10} + \sqrt[]{\dfrac{181}{20}})=0$

⇔\(\left[ \begin{array}{l}√5x-\dfrac{\sqrt[]{5}}{10} - \sqrt[]{\dfrac{181}{20}}=0\\√5x -\dfrac{\sqrt[]{5}}{10} + \sqrt[]{\dfrac{181}{20}}=0\end{array} \right.\) 

⇔\(\left[ \begin{array}{l}x=\dfrac{1+\sqrt[]{181}}{10}\\x=\dfrac{1-\sqrt[]{181}}{10}\end{array} \right.\) 

Vậy...

$d) -3x^2 -x+14=0$

$⇔ -3x^2 +6x-7x+14=0$

$⇔-3x(x-2) -7(x-2)=0$

$⇔(x-2)(-3x-7)=0$

⇔\(\left[ \begin{array}{l}x-2=0\\-3x-7=0\end{array} \right.\) 

⇔\(\left[ \begin{array}{l}x=2\\x=-\dfrac{7}{3}\end{array} \right.\) 

Vậy...

$e)-x+x+17=0$

$⇔ 0x +17=0$

$⇔ 0x=-17 (vô lí)

⇔ x vô nghiệm