Tìm tổng P + Q và hiệu P – Q của hai đa thức:

P = 4x²y² – 3xy³ + 5x³y – xy + 2x – 3;

Q = –4x²y² – 4xy³ – x³y + xy + y + 1.

Ta có:

U – 3x²y + 2xy² – 5y³ = 2xy² – xy + 1

Nên U = 2xy² – xy + 1 + 3x²y ‒ 2xy² + 5y³

           = (2xy² ‒ 2xy²) – xy + 3x²y + 5y³ + 1

           = ‒xy + 3x²y + 5y³ + 1.

Cách 1:

Ta có:

M + N ‒ P

= (3x³ – 5x²y + 5x – 3y) + (4xy – 4x + y) ‒ (3x³ + x²y + x + 1)

= 3x³ – 5x²y + 5x – 3y + 4xy – 4x + y ‒ 3x³ ‒ x²y ‒ x ‒ 1

= (3x³ ‒ 3x³) + (–5x²y ‒ x²y) + (5x – 4x ‒ x) + (– 3y + y) + 4xy ‒ 1

= ‒6x²y + 4xy ‒ 2y ‒1.

M – N – P

= (3x³ – 5x²y + 5x – 3y) ‒ (4xy – 4x + y) ‒ (3x³ + x²y + x + 1)

= 3x³ – 5x²y + 5x – 3y ‒ 4xy + 4x ‒ y ‒ 3x³ ‒ x²y ‒ x ‒ 1

= (3x³ ‒ 3x³) + (–5x²y ‒ x²y) + (5x + 4x ‒ x) + (–3y ‒ y) ‒ 4xy ‒ 1

= ‒6x²y + 8x ‒ 4xy ‒ 4y ‒1.

Cách 2:

Ta có:

M – P

= (3x³ – 5x²y + 5x – 3y) ‒ (3x³ + x²y + x + 1)

= 3x³ – 5x²y + 5x – 3y ‒ 3x³ ‒ x²y ‒ x ‒ 1

= (3x³ – 3x³) + (– 5x²y ‒ x²y) + (5x – x) – 3y – 1

= –6x²y + 4x – 3y – 1

Khi đó:

• M + N – P = M – P + N

= –6x²y + 4x – 3y – 1 + 4xy – 4x + y

= –6x²y + (4x – 4x) + (–3y + y) + 4xy – 1

= –6x²y – 2y + 4xy – 1.

• M – N – P = M – P – N

= –6x²y + 4x – 3y – 1 – (4xy – 4x + y)

= –6x²y + 4x – 3y – 1 – 4xy + 4x – y

= –6x²y + (4x + 4x) + (–3y – y) – 4xy – 1

= –6x²y + 8x – 4y – 4xy – 1.