Cho hai đa thức: M(x) = 2x^4 − 3x^3 − x + 7x^3 − 5x + 1; N(x) = −2x^3 + x^2 + 3x^4 + 5x − 2x^4 − 6 + x.

a) Đúng. Ta có:\(M\left( x \right) = 2{x^4} - 3{x^3} - x + 7{x^3} - 5x + 1\)

                                    \[ = 2{x^4} + \left( { - 3{x^3} + 7{x^3}} \right) + \left( { - x - 5x} \right) + 1\]

                                    \[ = 2{x^4} + 4{x^3} - 6x + 1\].

b) Đúng. Ta có: \(N\left( x \right) =  - 2{x^3} + {x^2} + 3{x^4} + 5x - 2{x^4} - 6 + x\)

                                    \( = \left( {3{x^4} - 2{x^4}} \right) - 2{x^3} + {x^2} + \left( {5x + x} \right) - 6\)

                                  \( = {x^4} - 2{x^3} + {x^2} + 6x - 6\).

c) Sai. Ta có \[M\left( 1 \right) = 2 \cdot {1^4} + 4 \cdot {1^3} - 6 \cdot 1 + 1 = 1\].

                   \(N\left( { - 1} \right) = {\left( { - 1} \right)^4} - 2 \cdot {\left( { - 1} \right)^3} + {\left( { - 1} \right)^2} + 6 \cdot \left( { - 1} \right) - 6 =  - 8\).

Do đó \(8M\left( 1 \right) + N\left( { - 1} \right) = 8 \cdot 1 + \left( { - 8} \right) = 0\).

d) Đúng. Ta có \(Q\left( x \right) = M\left( x \right) + N\left( x \right)\)

\(Q\left( x \right) = \left( {2{x^4} + 4{x^3} - 6x + 1} \right) + \left( {{x^4} - 2{x^3} + {x^2} + 6x - 6} \right)\)

         \( = 2{x^4} + 4{x^3} - 6x + 1 + {x^4} - 2{x^3} + {x^2} + 6x - 6\)

         \( = 3{x^4} + 2{x^3} + {x^2} - 5\).

Ta có \(Q\left( x \right) = 3{x^4} + 2{x^3} + 4\).