a, x^2 + 4x +4 = 16

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

<=> x^2 + 4x - 12 = 0

<=> x^2 + 4x - 12 = 0

<=> x^2 - 2x + 6x -12 = 0

<=> x(x - 2) + 6(x - 2) = 0

<=> (x - 2)(x + 6) = 0

<=> x -2 = 0 hoặc x + 6 =0

<=> x = 2 hoặc x = -6

b, x^3 + 3x^2 + 3x = 124

<=> x^3 + 3x^2 + 3x - 124 = 0

<=> x^3 - 4x^2 + 7x^2 - 28x + 31x - 124 = 0

<=>x^2(x - 4) + 7x(x - 4) + 31(x - 4) = 0

<=> (x - 4)(x^2 + 7x +31) = 0 (1)

Vì x^2 + 7x + 31 = (x^2 + 2.x.3,5 + 12,25) + 18,25 = (x + 3,5)^2 + 18,25 luôn lớn hơn 0 với mọi x

=> (1) <=> x - 4 = 0 

<=> x = 4

c, 2x + 8 = x^2

<=> -x^2 + 2x + 8 = 0

<=> x^2 - 2x - 8 = 0

<=> x^2 - 4x + 2x - 8 = 0

<=> x(x - 4) + 2(x - 4) = 0

<=> (x - 4)(x + 2) = 0

<=> x - 4 = 0 hoặc x + 2 = 0

<=> x = 4 hoặc x = -2

d, (x - 4)(x - 5) = 2

<=> x^2 - 5x - 4x + 20 = 2

<=> x^2 - 9x + 20 = 2

<=> x^2 - 9x + 20 - 2 = 0

<=> x^2 - 9x + 18 = 0

<=> x^2 - 3x - 6x + 18 = 0

<=> x(x - 3) - 6(x - 3) = 0

<=> (x - 3)(x - 6) = 0

<=> x - 3 = 0 hoặc x - 6 =0

<=> x = 3 hoặc x = 6

#NOCOPY
#NOPLAGIARISM

Đáp án+Giải thích các bước giải:

`a) x^2+4x+4=16`

`<=> x^2 + 2.x.2 +2^2=16`

`<=> (x+2)^2=16`

`<=> (x+2)^2=4^2`

`<=>`\(\left[ \begin{array}{l}x+2=4\\x+2=-4\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}x=2\\x=-6\end{array} \right.\) 

     Vậy `x∈{2; -6}`

`b) x^3+3x^2+3x =124`

`<=> x^3+3x^2+3x+1 =124+1`

`<=>  (x+1)^3 = 125`

`<=> (x+1)^3 = 5^3`

`<=> x+1=5`

`<=> x = 4`

   Vậy `x=4`

`c) 2x+8=x^2`

`<=> x^2-2x-8=0`

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

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

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

`<=>`\(\left[ \begin{array}{l}x-4=0\\x+2=0\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}x=4\\x=-2\end{array} \right.\) 

 Vậy `x ∈ {4; -2}`

`d) (x-4)(x-5)=2`

`<=> x^2-5x-4x+20-2=0`

`<=> x^2 -9x +18=0`

`<=> x^2-3x-6x+18=0`

`<=> (x-6)(x-3)=0`

`<=>`\(\left[ \begin{array}{l}x-6=0\\x-3=0\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}x=6\\x=3\end{array} \right.\) 

    Vậy `x∈{6; 3}`