a) =(x-y)⁵+(x-y)³=(x-y)³[(x-y)²+1]

b) =3³(y-2x)³:-9(y-2x)=-3(y-2x)²

c) =(x-y)² [3(x-y)³-2(x-y)²+3]:5(x-y)²=[3(x-y)³-2(x-y)²+3]/5

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

x⁴ - x³ + x² + 3x 

Ta có: x²-y²+6x+9= (x²+6x+9)-y²

                          = (x+3)² - y²

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

      Suy ra:  (x²-y²+6x+9):( (x+y+3) = (x+y+3)(x-y+3) : (x+y+3) = x-y+3

bạn chú ý nhé 
x^2-y^2+6x+9 = (x^2 +6x+9)-y^2(sử dụng hđt a^2+2ab+b^2=(a+b)^2 ) 
=(x+3)^2-y^2 (sử dụng hđt a^2 - b^2= (a+b)(a-b) 
=(x+3+y)(x+3-y) 
=>(x^2 -y^2+6x+9)/(x+y+3)=x-y+3 
chúc may mắn

a: \(=\dfrac{\left(x^4-y^4\right)^2}{x^2+y^2}=\left(x^2-y^2\right)^2\cdot\left(x^2+y^2\right)\)

b: \(=\dfrac{\left(4x+3\right)\left(16x^2-12x+9\right)}{16x^2-12x+9}=4x+3\)

b: Ta có: \(\left(4x^4-3x^3\right):\left(-x^3\right)+\left(15x^2+6x\right):3x=0\)

\(\Leftrightarrow-4x+3+5x+2=0\)

\(\Leftrightarrow x=-5\)

a. Làm gọn 1 chút xíu:

\(y=\left(x^{11}+2x^7-3x^5-6x\right)\left(3x^7+6x^2-2\right)\)

\(y'=\left(11x^{10}+14x^6-15x^4-6\right)\left(3x^7+6x^2-2\right)+\left(21x^6+12x\right)\left(x^{11}+2x^7-3x^5-6x\right)\)

b.

 \(y'=5\left(x^4-\dfrac{2}{3x}\right)^4\left(4x^3+\dfrac{2}{3x^2}\right)\Rightarrow y'\left(10\right)=5\left(10^4-\dfrac{2}{30}\right)^4\left(4.10^3+\dfrac{2}{300}\right)=?\)

c.

\(y'=\dfrac{7}{\left(x+1\right)^2}\Rightarrow y'\left(4\right)=\dfrac{7}{25}\)

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, = 3x^2y - 3xy + 6x^2y + 5xy - 9x^2y`

`= 2xy`.

Thay `x = 2/3; y = -3/4` vào BT:

`2 . 2/3 . -3/4 = -1.`

`b, x(x-2y) - y(y^2-2x)`

`= x^2 - 2xy - y^3 + 2xy`

`= x^2 - y^3`

Thay `x = 5; y =3` vào BT:

`= 5^2 - 3^3 = 25 - 27 = -2`

a) \(3x^2y-\left(3xy-6x^2y\right)+\left(5xy-9x^2y\right)\)

\(=3x^2y-3xy+6x^2y+5xy-9x^2y\)

\(=2xy\)

Thay \(x=\dfrac{2}{3},y=-\dfrac{3}{4}\) vào Bt ta có:

\(2\cdot\dfrac{2}{3}\cdot-\dfrac{3}{4}=-1\)

b) \(x\left(x-2y\right)-y\left(y^2-2x\right)\)

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

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

Thay \(x=5,y=3\) vào Bt ta có:
\(5^2-3^3=-3\)