\(a,=3x^3y^3-3x^2y^3+3x^2y^4+3xy^5\\ b,=\left(2x^3-6x^2+10x-3x^2+9x-15\right):\left(x^2-3x+5\right)\\ =\left[2x\left(x^2-3x+5\right)-3\left(x^2-3x+5\right)\right]:\left(x^2-3x+5\right)\\ =2x-3\\ c,=\left[x^2\left(x-3\right)+\left(x-3\right)\right]:\left(x-3\right)=x^2+1\)

a/ 5x²y (x²y– 4xy² + 7xy)

`=5x^4y^2-20x^3y^3+35x^3y^2`

b/ 3xy² (x²y³ + x 2y – xy² )

`=3x^3y^5+3x^3y^3-3x^2y^4`

c/ 3x(12x² + 4x – 5) + 2x(9x² – 6x + 7)

`=36x^3+12x^2-15x+18x^3-18x^2+14x`

`=54x^3-6x^2-x`

d/ 5x(2x² – 9x – 5) – 9x (x² - 7x – 4)

`=10x^3-45x^2-25x-9x^3+63x^2+36x`

`=x^3+18x^2+11x`

a: \(=15x^3-6x^2-3x\)

e: \(=x^2-12x+35\)

a) Ta có P = ( 4 x 2 − 1 ) ( 2 x + 1 ) − ( 2 x − 1 ) − ( 4 x 2 − 1 ) ( 2 x + 1 ) ( 2 x − 1 ) = 3 − 4 x 2  

b) Ta có  Q = 3 x ( x + 3 ) . ( x + 3 ) ( x − 3 ) − x = 9 − 3 x x + 3

a) = x^2 - 9 - (x^2 + 3x - 10)

= -3x + 1

b) = 3x + 1 - 3x + 19

= 20

a: \(\left(x+3\right)\left(x-3\right)-\left(x-2\right)\left(x+5\right)\)

\(=x^2-9-x^2-3x+10\)

\(=-3x+1\)

b: \(\dfrac{27x^3+1}{9x^2-3x+1}-\left(3x-19\right)\)

\(=3x+1-3x+19\)

=20

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

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

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

\(=\dfrac{-3\left(x-1\right)}{\left(x+1\right)^2}\cdot\dfrac{x+1}{6\left(x+1\right)\left(x-1\right)}\)

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

b) Bạn có thể viết kiểu latex được không ạ ?

`th1:`

`(x+1)(x^2-x+1):(x-3)(x^2+3x+9)`

`=(x^3+1^3):(x^3-3^3)`

`=(x^3+1):(x^3-27)`

`=(x^3+1)/(x^3-27)`

`=(x^3-27+28)/(x^3-27)`

`=1+28/(x^3-27)`

`**th2:`

`(x+1)(x^2-x+1)`

`=x^3+1^3=x^3+1`

`(x-3)(x^2+3x+9)`

`=x^3-3^3=x^3-27`

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

\(\left(x-3\right)\left(x^2+3x+9\right)=x^3-27\)