Giải SBT Toán 8 KNTT Bài 3: Phép cộng và phép trừ đa thức có đáp án

P + Q = 4x²y² – 3xy³ + 5x³y – xy + 2x – 3–4x²y² – 4xy³ – x³y + xy + y + 1

= (4x²y²–4x²y²) + (– 3xy³– 4xy³) + (5x³y– x³y) + (– xy + xy) + 2x + y + (–3 + 1)

= ‒7xy³ + 4x³y + 2x + y ‒ 2.

P ‒ Q = 4x²y² – 3xy³ + 5x³y – xy + 2x – 3 ‒ (–4x²y² – 4xy³ – x³y + xy + y + 1)

= 4x²y² – 3xy³ + 5x³y – xy + 2x – 3 + 4x²y² + 4xy³ + x³y ‒ xy ‒ y ‒ 1

= (4x²y²+4x²y²) + (– 3xy³+ 4xy³) + (5x³y+ x³y) + (– xy ‒ xy) + 2x ‒ y + (–3 ‒ 1)

= 8x²y² + xy³ + 6x³y ‒ 2xy + 2x ‒ y ‒ 4.

Ta có:

M + N

= 3x²y² – 0,8xy² + 2y² – 1–3x²y² – 0,2xy² + 2

= (3x²y²–3x²y²) + (– 0,8xy²– 0,2xy²) + 2y² + (–1 + 2)

= ‒xy² + 2y² + 1

Đa thức này có bậc 3, nhỏ hơn bậc của đa thức M (bậc 4).

Ta có:

U – 3x²y + 2xy² – 5y³ = 2xy² – xy + 1

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

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

           = ‒xy + 3x²y+ 5y³ + 1.

Do V + 4y³ – 2xy² + x²y – 9 = 4y³ – 3

NênV = 4y³ – 3 ‒ 4y³ + 2xy² – x²y + 9

           = (4y³ ‒ 4y³) + 2xy² ‒ x²y + (‒3 + 9)

           = 2xy² ‒ x²y + 6.

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.