giúp t bài 6 !

#linh

`1)` `8x(x - 5) - 3x + 15 = 0 `

`=> 8x^2 - 40x - 3x + 15 = 0 `

`=> 8x^2 - 43x + 15 = 0 `

`=> (8x - 3)(x - 5) = 0 `

`=> 8x - 3 = 0` hoặc `x - 5 = 0 `

`=> x = 3/8` hoặc `x = 5`

Vậy `x in {3/8;5}`

``

`2)` `2x^3 - 50x = 0`

`=> 2x(x^2 - 25) = 0`

`=> 2x(x - 5)(x + 5) = 0`

`=> 2x = 0` hoặc `x - 5 = 0` hoặc `x + 5 = 0`

`=> x = 0` hoặc `x = 5` hoặc `x = -5`

Vậy `x in {0;5;-5}`

``

`3)` `2x^2 - 2x = (x - 1)^2 `

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

`=> x^2 - 1 = 0 `

`=> (x - 1)(x + 1) = 0 `

`=> x - 1 = 0` hoặc `x + 1 = 0`

`=> x = 1` hoặc `x = -1`

Vậy `x in {1;-1}`

``

`4)` `9x^2 - 16 + (3x + 4)^2 = 0`

`=> (3x)^2 - 4^2 + (3x + 4)^2 = 0`

`=> (3x - 4)(3x + 4) + (3x + 4)^2 = 0`

`=> (3x + 4)(3x - 4 + 3x + 4) = 0`

`=> 6x(3x + 4) = 0`

`=> 6x = 0` hoặc `3x + 4 = 0`

`=> x = 0` hoặc `x = -4/3`

Vậy `x in {0;-4/3}`

``

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

`=> (2x - 3)^2 - (x + 5)^2 = 0 `

`=> [(2x - 3) - (x + 5)][(2x - 3) + (x + 5)] = 0`

`=> (x - 8)(3x + 2) = 0`

`=> x - 8 = 0` hoặc `3x + 2 = 0`

`=> x = 8` hoặc `x = -2/3`

Vậy `x in {8;-2/3}`

``

`6)` `4x^2 - 25 - (2x - 5)(2x + 7) = 0`

`=> (2x)^2 - 5^2 - (2x - 5)(2x + 7) = 0`

`=> (2x - 5)(2x + 5) - (2x - 5)(2x + 7) = 0`

`=> (2x - 5)(2x + 5 - 2x - 7) = 0`

`=> -2(2x - 5) = 0 `

`=> 2x - 5 = 0`

`=> x = 5/2`

Vậy `x = 5/2`