Bài tập 1.17 trang 11 SBT Toán 8 Tập 1 Kết nối tri thức - KNTT

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.

-- Mod Toán 8 HỌC247