Tìm x , biết: a) ( x + 1 ) ( x^2 − x + 1 ) − x^3 + 2x = 0 ;

Hướng dẫn giải

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

\[{x^3} + 1 - {x^3} + 2x = 0\]

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

 \(2x + 1 = 0\)

\(2x =  - 1\)

\(x =  - \frac{1}{2}.\)

Vậy \(x =  - \frac{1}{2}.\)

b) \[\left( {5 - 3x} \right)\left( {2{x^2} + 3x + 3} \right) + 5{x^2}\left( {3x - 5} \right) =  - 15{x^2} + 9{x^3}\]

\[\left( {5 - 3x} \right)\left( {2{x^2} + 3x + 3} \right) - 5{x^2}\left( {5 - 3x} \right) =  - 3{x^2}\left( {5 - 3x} \right)\]

\[\left( {5 - 3x} \right)\left( {2{x^2} + 3x + 3} \right) - 5{x^2}\left( {5 - 3x} \right) + 3{x^2}\left( {5 - 3x} \right) = 0\]

\[\left( {5 - 3x} \right)\left( {2{x^2} + 3x + 3 - 5{x^2} + 3{x^2}} \right) = 0\]

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

Suy ra \[5 - 3x = 0\] hoặc \[3x + 3 = 0\]

 \[3x = 5\] hoặc \[3x =  - 3\]

 \[x = \frac{5}{3}\] hoặc \[x =  - 1\]

Vậy \[x \in \left\{ {\frac{5}{3}; - 1} \right\}.\]