Đề bài : Tìm giá trị của x

a. x² - ( x + 4 ) × ( x + 3 ) = 26

b. ( 3x - 1 ) × ( x + 3 ) - ( x + 2 ) = 1

c. 1/2 x² - ( x - 4 ) × ( 1/2 x + 3 ) = 12

`a) x^2 - (x+4) xx (x+3)=26`

`=> x^2 - x^2 - 7x - 12 = 26`

`=> -7x=26+12`

`=> -7x=38`

`=> x=-38/7`

Vậy `x=-38/7`

`b) (3x-1) xx (x+3) - (x+2)=1`

`=> 3x^2 + 8x - 3 - x - 2-1=0`

`=> 3x^2 + 7x - 6=0` 

`=> 3x^2 - 2x + 9x - 6 =0`

`=> (3x^2 - 2x) + (9x-6)=0`

`=> x(3x - 2) + 3(3x-2)=0`

`=> (x+3)(3x-2)=0`

`=> x+3=0` hoặc `3x-2=0`

`=> x=-3` hoặc `x=2/3`

Vậy `x in {-3;2/3}`

`c) 1/2 x^2 - (x-4) xx (1/2 x +3)=12`

`=> 1/2 x^2 - (1/2x^2 + x - 12)=12`

`=> 1/2 x^2 - 1/2 x^2 - x + 12=12`

`=> -x=0`

`=> x=0`

Vậy `x=0`