Question

Tìm đa thức P và đa thức Q biết :

a) \(P+\left(x^2-2y^2\right)=x^2-y^2+3y^2-1\)

b) \(Q-\left(5x^2-xyz\right)=xy+2x^2-3xyz+5\)

Answer 1

a) P + (x² - 2y²) = x² - y² + 3y² - 1

⇔ P = (x² - y² + 3y² - 1) - (x² - 2y²)

⇔ P = x² + 2y² - 1 - x² + 2y²

⇔ P = 4y² - 1

b) Q - (5x² - xyz) = xy + 2x² - 3xyz + 5

⇔ Q = (xy + 2x² - 3xyz + 5) + (5x² - xyz)

⇔ Q = xy + 2x² - 3xyz + 5 + 5x² - xyz

⇔ Q = xy + 7x² - 4xyz + 5

Answer 2

a) P + (x² – 2y²) = x² – y² + 3y² – 1

P = (x² – y² + 3y² – 1) - (x² – 2y²)

P = x² – y² + 3y² – 1 - x² + 2y²

P = x² – x² – y² + 3y² + 2y² – 1

P = 4y² – 1.

Vậy P = 4y² – 1.

b) Q – (5x² – xyz) = xy + 2x² – 3xyz + 5

Q = (xy + 2x² – 3xyz + 5) + (5x² – xyz)

Q = xy + 2x² – 3xyz + 5 + 5x² – xyz

Q = 7x² – 4xyz + xy + 5

Vậy Q = 7x² – 4xyz + xy + 5.

Answer 3

a) P + (x2 – 2y2) = x2 – y2 + 3y2 – 1

P = (x2 – y2 + 3y2 – 1) – (x2 – 2y2)

P = x2 – y2 + 3y2 – 1 – x2 + 2y2

P = x2 – x2 – y2 + 3y2 + 2y2 – 1

P = 4y2 – 1

b) Q – (5x2 – xyz) = xy + 2x2 – 3xyz + 5

Q = (xy + 2x2 – 3xyz + 5) + (5x2 – xyz)

Q = xy + 2x2 – 3xyz + 5 + 5x2 – xyz

Q = 7x2– 4xyz + xy + 5