PHẦN II. TỰ LUẬN (8,0 điểm)

(1,0 điểm) Tìm \(x,\) biết:

a) \({\left( {x + 3} \right)^2} + \left( {4 - x} \right)\left( {x + 4} \right) = 10.\)                                 

b) \({\left( {{x^2} - 9} \right)^2} - {\left( {x - 3} \right)^2} = 0.\)

a) \({\left( {x + 3} \right)^2} + \left( {4 - x} \right)\left( {x + 4} \right) = 10\)

\({x^2} + 6x + 9 + 16 - {x^2} = 10\)

\[\left( {{x^2} - {x^2}} \right) + 6x = 10 - 9 - 16\]

\[6x = - 15\]

\(x = - \frac{5}{2}.\)

Vậy \(x = - \frac{5}{2}.\)

b) \({\left( {{x^2} - 9} \right)^2} - {\left( {x - 3} \right)^2} = 0\)

\({\left[ {\left( {x - 3} \right)\left( {x + 3} \right)} \right]^2} - {\left( {x - 3} \right)^2} = 0\)

\({\left( {x - 3} \right)^2}{\left( {x + 3} \right)^2} - {\left( {x - 3} \right)^2} = 0\)

\({\left( {x - 3} \right)^2}\left[ {{{\left( {x + 3} \right)}^2} - 1} \right] = 0\)

\({\left( {x - 3} \right)^2}\left[ {\left( {x + 3 - 1} \right)\left( {x + 3 + 1} \right)} \right] = 0\)

\({\left( {x - 3} \right)^2}\left( {x + 2} \right)\left( {x + 4} \right) = 0\)

Suy ra \({\left( {x - 3} \right)^2} = 0\) hoặc \(x + 2 = 0\) hoặc \(x + 4 = 0\)

\(x - 3 = 0\) hoặc \(x = - 2\) hoặc \(x = - 4\)

\(x = 3\) hoặc \(x = - 2\) hoặc \(x = - 4\)

Vậy \[x \in \left\{ {3\,;\,\, - 2\,;\,\, - 4} \right\}.\]