Đáp án:

`a) 6x - 2`

`b) - 8x^2 - 7x + 17`

`c) {- 1}/{x - 3}`

Giải thích các bước giải:

 `a)`

`6x(x + 1) - 2(3x^2  +1)`

`= 6x^2 + 6x - 6x^2 - 2`

`= (6x^2 - 6x62) + 6x - 2`

`= 6x - 2`

`b)`

`(x + 4)^2 - (2x - 1)(2x + 1) - 5x(x + 3)`

`= (x^2 + 2.x.4 + 4^2) - [(2x)^2 - 1^2] - 5x^2 - 15x`

`= (x^2 + 8x + 16) - (4x^2  - 1) - 5x^2 - 15x`

`= x^2 + 8x + 16 - 4x^2 + 1 - 5x^2 - 15x`

`= (x^2 - 4x^2 - 5x^2) + (8x - 15x) + (16 + 1)`

`= - 8x^2 - 7x + 17`

`c)`

`{x - 3}/{x + 3} - {x + 3}/{x - 3} + {11x - 3}/{x^2 - 9}`

`= {x - 3}/{x + 3} - {x + 3}/{x - 3} + {11x - 3}/{x^2 - 3^2}`

`= {x - 3}/{x + 3} - {x + 3}/{x - 3} + {11x - 3}/{(x - 3)(x + 3)}`

`= {(x - 3)(x - 3)}/{(x - 3)(x + 3)} - {(x + 3)(x + 3)}/{(x - 3)(x + 3)} + {11x - 3}/{(x - 3)(x + 3)}`

`= {(x - 3)^2}/{(x - 3)(x + 3)} - {(x + 3)^2}/{(x - 3)(x + 3)} + {11x - 3}/{(x - 3)(x + 3)}`

`= {x^2 - 6x + 9}/{(x - 3)(x + 3)} - {x^2 + 6x + 9}/{(x - 3)(x + 3)} + {11x - 3}/{(x - 3)(x + 3)}`

`= {x^2 - 6x + 9- (x^2 + 6x +  9) + 11x - 3}/{(x - 3)(x + 3)}`

`= {x^2 - 6x + 9- x^2 - 6x - 9 + 11x - 3}/{(x - 3)(x + 3)}`

`= {- x - 3}/{(x - 3)(x + 3)}`

`= {- (x + 3)}/{(x - 3)(x + 3)}`

`= {- 1}/{x - 3}`

Bài 1:

a) 6x(x + 1) - 2(3x² + 1)

= 6x² + 6x - 6x² - 2

= 6x - 2

b) (x + 4)² - (2x - 1)(2x + 1) - 5x(x + 3)

= x² + 8x + 16 - [(2x)² - 1] - 5x² - 15

= x² + 8x + 16 - 4x² + 1 - 5x² - 15

= -8x² + 8x + 2

(nếu nó là + (2x + 1)(2x - 1) thì số đẹp hơn đấy!! :)))

c) $\frac{x - 3}{x + 3}$ - $\frac{x + 3}{x - 3}$ + $\frac{11x - 3}{x² - 9}$

= $\frac{x - 3}{x + 3}$ - $\frac{x + 3}{x - 3}$ + $\frac{11x - 3}{(x - 3)(x + 3)}$

= $\frac{(x - 3)²}{(x + 3)(x - 3)}$ - $\frac{(x + 3)²}{(x - 3)(x + 3)}$ + $\frac{11x - 3}{(x - 3)(x + 3)}$

= $\frac{(x² -6x + 9) - (x² + 6x + 9) + (11x - 3)}{(x + 3)(x - 3)}$

= $\frac{(-x) - 3}{(x + 3)(x - 3)}$

= $\frac{- (x + 3)}{(x + 3)(x - 3)}$ = $\frac{-1}{(x - 3)}$ = $\frac{1}{(3 - x)}$