Bài 12: Chia đa thức một biến đã sắp xếp

a) (x³ – 7x + 3 – x²) : (x – 3)

b) (2x⁴ – 3x² – 3x² – 2 + 6x) : (x² – 2)

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a) (x² + 2xy + y²) : (x + y) = (x + y)² : (x + y) = x + y.

b) (125x³ + 1) : (5x + 1) = [(5x)³ + 1] : (5x + 1)

= (5x)² – 5x + 1 = 25x² – 5x + 1.

c) (x² – 2xy + y²) : (y – x) = (x – y)² : [-(x – y)] = - (x – y) = y – x

Hoặc (x² – 2xy + y²) : (y – x) = (y² – 2xy + x²) : (y – x)

= (y – x)² : (y – x) = y - x.

Vậy 3x⁴ + x³ + 6x – 5 = (x²+ 1)(3x² + x – 3) + 5x - 2

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a) (25x⁵ – 5x⁴ + 10x²) : 5x² = (25x⁵ : 5x² ) - (5x⁴ : 5x² ) + (10x² : 5x² )

= 5x³ – x² + 2

b) (15x³y² – 6x²y – 3x²y²) : 6x²y

= (15x³y² : 6x²y) + (– 6x²y : 6x²y) + (– 3x²y² : 6x²y)

= \(\dfrac{15}{6}\)xy - 1 - \(\dfrac{3}{6}\)y = \(\dfrac{5}{2}\)xy - \(\dfrac{1}{2}\)y - 1.

a) A chia hết cho B vì x⁴, x³, x² đều chia hết cho x²

b) A chia hết cho B, vì x²– 2x + 1 = (1 – x)², chia hết cho 1 - x

Ta có:\(\left(2x^4+x^3-3x^2+5x-2\right):\left(x^2-x+1\right)\)

= \(\left(2x^4-2x^3+2x^2+3x^3-3x^2+3x-2x^2+2x-2\right):\left(x^2-x+1\right)\)

=\(\left(\left(2x^4-2x^3+2x^2\right):\left(x^2-x+1\right)\right)\)+\(\left(\left(3x^3-3x^2+3x\right):\left(x^2-x+1\right)\right)\)+\(\left(\left(-2x^2+2x-2\right):\left(x^2-x+1\right)\right)\)

= \(2x^2.\left(x^2-x+1\right):\left(x^2-x+1\right)\)+\(3x.\left(x^2-x+1\right):\left(x^2-x+1\right)\)\(-2\left(x^2-x+1\right):\left(x^2-x+1\right)\)

= \(2x^2+3x-2\)

a) (4x² – 9y²) : (2x – 3y) = [(2x)² – (3y)²] : (2x – 3y) = 2x + 3y;

b) (27x³ – 1) : (3x – 1) = [(3x)³ – 1] : (3x – 1) = (3x)² + 3x + 1 = 9x² + 3x + 1

c) (8x³ + 1) : (4x² – 2x + 1) = [(2x)³ + 1] : (4x² – 2x + 1)

= (2x + 1)[(2x)² – 2x + 1] : (4x² – 2x + 1)

= (2x + 1)(4x² – 2x + 1) : (4x² – 2x + 1) = 2x + 1

d) (x² – 3x + xy -3y) : (x + y)

= [(x² + xy) – (3x + 3y)] : (x + y)

= [x(x + y) – 3(x + y)] : (x + y)

= (x + y)(x – 3) : (x + y)

= x – 3.

Đề phép chia hết thì dư a - 30 phải bằng 0 tức là

a - 30 = 0 => a = 30

Vậy a = 30.

a: \(=\dfrac{6x^2+15x-2x-5}{2x+5}=3x-1\)

b: \(=\dfrac{x^2\left(x+3\right)+\left(x-3\right)}{x-3}=x^2+1\)

c: \(=\dfrac{2x^4-6x^2+x^3-3x+x^2-3}{x^2-3}=2x^2+x+1\)

a: \(=\dfrac{x^4-6x^3+12x^2-14x+3}{x^2-4x+1}\)

\(=\dfrac{x^4-4x^3+x^2-2x^3+8x^2-2x+3x^2-12x+3}{x^2-4x+1}\)

\(=x^2-2x+3\)

b: \(=\dfrac{x^5-3x^4+5x^3-x^2+3x-5}{x^2-3x+5}=x^2-1\)

c: \(=\dfrac{2x^4-5x^3+2x^2+2x-1}{x^2-x-1}\)

\(=\dfrac{2x^4-2x^3-2x^2-3x^3+3x^2+3x+x^2-x-1}{x^2-x-1}\)

\(=2x^2-3x+1\)