Đáp án + Giải thích các bước giải:

`a)`

`8x(x - 2017) - 2x + 4034 = 0`

`<=> 8x(x - 2017) - 2(x - 2017) = 0`

`<=> (8x - 2)(x - 2017) = 0`

`<=> [(8x -2 =0),(x - 2017 = 0):}`

`<=> [(8x = 2),(x = 2017):}`

`<=> [(x = 1/4),(x = 2017):}`

Vậy `S = {1/4 ; 2017}`

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`b)`

`4 - x = 2(x - 4)^2`

`<=> 4 - x - 2(x^2 - 8x + 16) = 0`

`<=> 4 - x - 2x^2 + 16x - 32 = 0`

`<=> -2x^2 + 15x - 28 = 0`

`<=> -(2x^2 - 15x + 28) = 0`

`<=> 2x^2 - 15x + 28 = 0`

`<=> 2x^2 - 8x - 7x + 28 = 0`

`<=> 2x(x - 4) - 7(x - 4) = 0`

`<=> (2x - 7)(x - 4) = 0`

`<=> [(2x - 7 = 0),(x - 4 = 0):}`

`<=> [(2x = 7),(x = 4):}`

`<=> [(x = 7/2),(x = 4):}`

Vậy `S = {7/2 ; 4}`

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`c)`

`(x^2 + 1)(x - 2) + 2x = 4`

`<=> (x^2 + 1)(x - 2) + 2x - 4 = 0`

`<=> (x^2 + 1)(x - 2) + 2(x - 2) = 0`

`<=> (x^2 + 1 + 2)(x - 2) = 0`

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

`<=> [(x^2 + 3 = 0  (mà  x^2 >= 0 AA x)),(x - 2 = 0):}`

`<=> [(x^2 + 3 = 0  (vô  lí)),(x = 2):}`

Vậy `S = {2}`

Giải thích các bước giải:

`a) 8x(x-2017) - 2x + 4034 = 0`

`⇔ 8x( x-2017 ) + 4034 - 2x = 0`

`⇔ 8x( x-2017) + 2( 2017 - x ) = 0`

`⇔ 8x( x-2017) - 2( x-2017 ) = 0`

`⇔ ( 8x-2 )( x-2017 ) = 0`

`⇔ 8x-2=0` hoặc `x-2017=0`

`⇔ 8x = 2` hoặc `x=2017`

`⇔ x = 1/4` hoặc `x=2017`

Vậy `S={1/4;2017}`

`b) 4 - x = 2(x-4)^2`

`⇔ 4-x = 2( 4-x )^2`

`⇔ 2( 4-x )^2 - ( 4-x ) = 0`

`⇔ [2( 4-x ) - 1 ]( 4-x ) = 0`

`⇔ ( 8 - 2x - 1 )( 4-x ) = 0`

`⇔ ( 7 - 2x )( 4-x ) = 0`

`⇔ 7-2x=0` hoặc `4-x=0`

`⇔ 2x=7` hoặc `x=4`

`⇔ x=7/2` hoặc `x=4`

Vậy `S={7/2;4}`

`c) (x^2+1)(x-2)+2x=4`

`⇔ ( x^2 + 1 )( x-2 ) + 2x-4 =0`

`⇔ ( x^2 + 1 )( x-2 ) + 2( x-2 ) = 0`

`⇔ ( x^2 + 1 + 2 )( x-2 ) = 0`

`⇔ ( x^2 + 3 )( x-2 ) = 0`

Do `x^2 + 3 ≥ 3 \ne 0 AAx`

`⇒ x-2=0`

`⇔ x=2`

Vậy `S={2}`