Tìm x , biết: a) ( x + 3 )^2 + ( 4 − x ) ( x + 4 ) = 10. b) x^2 − 2x = 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) \({x^2} - 2x = 0\)

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

Suy ra \(x = 0\) hoặc \(x - 2 = 0\)

\(x = 0\) hoặc \(x = 2\)

Vậy \(x \in \left\{ {0;2} \right\}.\)

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

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

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

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

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

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

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 \(x \in \left\{ {3; - 2; - 4} \right\}.\)