Tìm x, biết: 1)3(x^2) + 15x = 0

Hướng dẫn giải

1) \(3{x^2} + 15x = 0\)

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

\[x = 0\] hoặc \[x + 5 = 0\]

\[x = 0\] hoặc \[x =  - 5.\]

Vậy \[x \in \left\{ {0\,;\,\, - 5} \right\}.\]

2) \(\left( {x + 2} \right)\left( {{x^2} - 2x + 4} \right) - x\left( {{x^2} - 3} \right) = 14\)

\({x^3} + 8 - {x^3} + 3x = 14\)

\(3x + 8 = 14\)

\(3x = 6\)

\(x = 2.\)

Vậy \(x = 2.\)

3) \[{x^2} - \left( {x + 2} \right)\left( {3x - 7} \right) = 4\]

\[{x^2} - 4 - \left( {x + 2} \right)\left( {3x - 7} \right) = 0\]
\[\left( {x + 2} \right)\left( {x - 2} \right) - \left( {x + 2} \right)\left( {3x - 7} \right) = 0\]

\[\left( {x + 2} \right)\left( {x - 2 - 3x + 7} \right) = 0\]

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

\[x + 2 = 0\] hoặc \[ - 2x + 5 = 0\]

\[x =  - 2\]hoặc \[x = \frac{5}{2}.\]

Vậy \[x \in \left\{ { - 2\,;\,\,\frac{5}{2}} \right\}.\]