\(a)\) \({x^2} - 2x - 3 = 0\)
\( \Leftrightarrow {x^2} - 2x + 1 - 4 = 0 \)
\(\Leftrightarrow {\left( {x - 1} \right)^2} - {2^2} = 0 \)
\( \Leftrightarrow \left( {x - 1 + 2} \right)\left( {x - 1 - 2} \right) = 0 \)
\(\Leftrightarrow \left( {x + 1} \right)\left( {x - 3} \right)=0 \)
Suy ra \( x + 1 = 0\) hoặc \(x - 3 = 0\)
+) Với \(x+1=0\Leftrightarrow x=-1\)
+) Với \(x-3=0\Leftrightarrow x=3\)
Vậy \(x = - 1;\) \(x = 3\)
\(b)\) \(2{x^2} + 5x - 3 = 0\)
\(\Leftrightarrow 2{x^2} + 6x - x - 3 = 0 \)\(\Leftrightarrow 2x\left( {x + 3} \right) - \left( {x + 3} \right) = 0 \)\(\Leftrightarrow \left( {x + 3} \right)\left( {2x - 1} \right) = 0 \)
Suy ra \( x + 3 = 0\) hoặc \(2x - 1 = 0\)
+) Với \(x+3=0\Leftrightarrow x=-3\)
+) Với \(2x-1=0\Leftrightarrow 2x=1\)\(\Leftrightarrow x=\dfrac {1}{2}\)
Vậy \(x = - 3;\) \(x =\displaystyle{1 \over 2}\)
