Tìm x, biết: a) x{x + 3} --{x^2} = 9  

a) \[x\left( {x + 3} \right)--{x^2} = 9\]

\[{x^2} + 3x--{x^2} = 9\]

\[3x = 9\]

\[x = 3\].

Vậy \[x = 3\].

b) \[{\left( {x - 5} \right)^2} = \;4{x^2}\]

\[{\left( {x - 5} \right)^2} - {\left( {\;2x} \right)^2} = 0\]

\[\left( {x--5--2x} \right)\left( {x--5 + 2x} \right) = 0\]

\[\left( { - x--5} \right)\left( {3x--5} \right) = 0\]

\[ - x--5 = 0\] hoặc \[3x--5 = 0\]

\[x = - 5\] hoặc \[x = \frac{5}{3}\].

Vậy\[x \in \left\{ { - 5\,;\,\,\frac{5}{3}} \right\}\].