Question

Làm tính chia :

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

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

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

Answer 1

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

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

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

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

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

$= x + 3 - y$

$= x - y + 3$

Answer 2

(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

Answer 3

c)

=

=

=

=

= x + 3 - y

= x - y + 3

Answer 4

a) \(\dfrac{6x^3-7x^2-x+2}{2x+1}=\dfrac{6x^3-10x^2+4x+3x^2-5x+2}{2x+1}\)

\(=\dfrac{2x\left(3x^2-5x+2\right)+\left(3x^2-5x+2\right)}{2x+1}=\dfrac{\left(2x+1\right)\left(3x^2-5x+2\right)}{2x+1}\)

\(=3x^2-5x+2\)

b) \(\dfrac{x^4-x^3+x^2+3x}{x^2-2x+3}=\dfrac{x\left(x^3-x^2+x+3\right)}{x^2-2x+3}\)

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

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

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

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

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

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