Question

Cộng, trừ đa thức một biến:

M(x) = \(x^4+5x^3-x^2+x-0,5\)

N(x) = \(3x^4-5x^2-x-2,5\)

Đề bài: a) M(x)+N(x)

b) M(x)-N(x)

Answer 1

a) \(M\left(x\right)+N\left(x\right)=\left(x^4+5x^3-x^2+x-0,5\right)+\left(3x^4-5x^2-x-2,5\right)\)

\(=\left(x^4+3x^4\right)+5x^3-\left(x^2+5x^2\right)+\left(x-x\right)-\left(0,5+2,5\right)\)

\(=4x^4+5x^3-6x^2-3\)

b) \(M\left(x\right)-N\left(x\right)=\left(x^4+5x^3-x^2+x-0,5\right)-\left(3x^4-5x^2-x-2,5\right)\)

\(=x^4+5x^3-x^2+x-0,5-3x^4+5x^2+x+2,5\)

\(=\left(x^4-3x^4\right)+5x^3-\left(x^2-5x^2\right)+\left(x+x\right)-\left(0,5-2,5\right)\)

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

Answer 2

a) \(4x^4+5x^3-6x^2-3\)

b) \(-2x^4+5x^3-6x^2-3\)