a) \(\left( {6{x^3} - 7{x^2} - x + 2} \right):\left( {2x + 1} \right)\)

b) $(x^4-x^3+x^2+3x):(x^2-2x+3)$

c) \(\left( {{x^2} + {y^2} + 6x + 9} \right):\left( {x + y + 3} \right)\)

\(=\left( {{x^2} + 6x + 9 - {y^2}} \right)\left( {x + y + 3} \right)\)

\(=\left[ {\left( {{x^2} + 2x.3 + {3^2}} \right) - {y^2}} \right]:\left( {x + y + 3} \right)\)

\(=\left[ {{{\left( {x + 3} \right)}^2} - {y^2}} \right]:\left( {x + y + 3} \right)\)

\(=\left( {x + 3 - y} \right)\left( {x + 3 + y} \right):\left( {x + y + 3} \right)\)

$= x + 3 - y$

$= x - y + 3$

(6x³ - 7x² - x + 2) : (2x + 1)

= (6x³ + 3x² - 10x² - 5x + 4x + 2) : (2x + 1)

= [(6x³ + 3x²) - (10x² + 5x) + (4x + 2)] : (2x + 1)

= [3x²(2x + 1) - 5x(2x + 1) + 2(2x + 1)] : (2x + 1)

= (3x² - 5x + 2)(2x + 1) : (2x + 1)

= 3x² - 5x + 2

(x⁴ - x³ + x² + 3x) : (x² - 2x + 3)

= (x⁴ + x³ - 2x³ - 2x² + 3x² + 3x) : (x² - 2x + 3)

= [(x⁴ + x³) - (2x³ + 2x²) + (3x² + 3x)] : (x² - 2x + 3)

= [x³(x + 1) - 2x²(x + 1) + 3x(x + 1)] : (x² - 2x + 3)

= (x³ - 2x² + 3x)(x + 1) : (x² - 2x + 3)

= x(x² - 2x + 3)(x + 1): (x² - 2x + 3)

= x(x + 1)

= x² + x

(x² - y² + 6x + 9) : (x + y + 3)

= [(x² + 6x + 9) - y²] : (x + y + 3)

= [(x + 3)² - y²] : (x + y + 3)

= (x + 3 + y)(x + 3 - y) : (x + y + 3)

= (x + y + 3)(x - y + 3) : (x + y + 3)

= x - y + 3

CHÚC BN HOK TỐT

c)

=

=

=

=

= x + 3 - y

= x - y + 3

\(=\left(6x^3-7x^2-x+2\right).\frac{1}{2x+1}\)

a: \(=-2x^2\cdot3x+2x^2\cdot4X^3-2x^2\cdot7+2x^2\cdot x^2\)

\(=8x^5+2x^4-6x^3-14x^2\)

b: \(=2x^3-3x^2-5x+6x^2-9x-15\)

\(=2x^3+3x^2-14x-15\)

c: \(=\dfrac{-6x^5}{3x^3}+\dfrac{7x^4}{3x^3}-\dfrac{6x^3}{3x^3}=-2x^2+\dfrac{7}{3}x-2\)

d: \(=\dfrac{\left(3x-2\right)\left(3x+2\right)}{3x+2}=3x-2\)

e: \(=\dfrac{2x^4-8x^3-6x^2-5x^3+20x^2+15x+x^2-4x-3}{x^2-4x-3}\)

=2x^2-5x+1

1: \(\dfrac{6x^3-7x^2-x+2}{2x+2}\)

\(=\dfrac{6x^3+6x^2-13x^2-13x+12x+12-10}{2x+2}\)

\(=\dfrac{3x^2\left(2x+2\right)-\dfrac{13}{2}x\left(2x+2\right)+6\left(2x+2\right)-10}{2x+2}\)

\(=3x^2-\dfrac{13}{2}x+6-\dfrac{5}{x+1}\)

2: \(\dfrac{x^2-y^2+6x-9}{x+y+3}\)

\(=\dfrac{x^2-\left(y-3\right)^2}{x+y+3}\)

\(=\dfrac{\left(x-y+3\right)\left(x+y-3\right)}{x+y+3}\)

\(a,=\left(6x^3+3x^2-10x^2-5x+4x+2\right):\left(2x+1\right)\\ =\left[3x^2\left(2x+1\right)-5x\left(2x+1\right)+2\left(2x+1\right)\right]:\left(2x+1\right)\\ =3x^2-5x+2\\ b,Sửa:\left(2x^3-21x^2+67x-60\right):\left(x-5\right)\\ =\left(2x^3-10x^2-11x^2+55x+12x-60\right):\left(x-5\right)\\ =\left[2x^2\left(x-5\right)-11x\left(x-5\right)+12\left(x-5\right)\right]:\left(x-5\right)\\ =2x^2-11x+12\)

a) (x³ – 7x + 3 – x²) : (x – 3)

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

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