Cho hai đa thức: P ( x ) = 2 x^2 − 3 x^3 + x^2 + 3 x^3 − x − 1 − 3 x Q ( x ) = − 3 x^2 + 2 x^3 − x − 2 x^3 − 3 x − 2 (a) Thu gọn và sắp xếp hai đa thức P ( x ) ; Q ( x ) theo lũy thừa gi

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

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

\[ = 3{x^2} - 4x - 1\]

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

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

\[ = - 3{x^2} - 4x - 2\]

c) Ta có: \[g\left( x \right) = P\left( x \right) - Q\left( x \right)\]

Suy ra \(g\left( x \right) = \left( {3{x^2} - 4x - 1} \right) - \left( { - 3{x^2} - 4x - 2} \right)\)

\[ = 3{x^2} - 4x - 1 + 3{x^2} + 4x + 2\]

\( = 6{x^2} + 1\)

Do đó \[g\left( x \right) - \left( {6x + 1} \right) = 6{x^2} + 1 - 6x - 1 = 6{x^2} - 6x\]

Ta có: \[g\left( x \right) - 6x + 1 = 0\]

\[6{x^2} - 6x = 0\]

\[6x\left( {x - 1} \right) = 0\]

\(x = 0\) hoặc \(x = 1\)

Vậy \[x \in \left\{ {0;1} \right\}\] thì \[g\left( x \right) - \left( {6x + 1} \right) = 0\].