Question

Answer 1

\(a.M+N\\ =\left(5x^3-4x^3y+3y^2\right)+\left(-2x^3+4x^3y-y^2\right)\\ =5x^3-4x^3y+3y^2-2x^3+4x^3y-y^2\\ =\left(5x^3-2x^3\right)+\left(-4x^3y+4x^3y\right)+\left(3y^2-y^2\right)\\ =3x^3+2y^2\\ M-N\\ =\left(5x^3-4x^3y+3y^2\right)-\left(-2x^3+4x^3y-y^2\right)\\ =5x^3-4x^3y+3y^2+2x^3-4x^3y+y^2\\ =\left(5x^3+2x^3\right)+\left(-4x^3y-4x^3y\right)+\left(3y^2+y^2\right)\\ =7x^3-8x^3y+4y^2\) 

b) Bậc của M là: 3 + 1 = 4 

Khi x = 0 ta có: 

\(M=5\cdot0^3-4\cdot0^3y+3y^2=3y^2\)

Mà: M = 27 

=> \(3y^2=27=>y^2=9=>y=\pm3\)