Bài 5: Tìm x, biết: a) 8x (x-2017) -2x+4034 =0 b) $\frac{x}{2}$ + $\frac{$x^{2}$ }{8}$ =0 c) 4-x =2 $(x-4)^{2}$ d) ( $x^{2}$ +1) (x-2) +2x =4

`#Sad`

`a)`

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

`⇔ 8x(x-2017)-2(x-2017) = 0`

`⇔ (8x-2)(x-2017) = 0`

`⇔` \(\left[ \begin{array}{l}8x-2=0\\x-2017=0\end{array} \right.\) 

`⇔` \(\left[ \begin{array}{l}8x=2\\x=2017\end{array} \right.\) 

`⇔` \(\left[ \begin{array}{l}x=\dfrac{1}{4}\\x=2017\end{array} \right.\) 

`\text{Vậy S=}` `{1/4; 2017}`

`b)`

`x/2+(x^2)/8 = 0`

`⇔ (4x)/8+(x^2)/8 = 0`

`⇔ 4x+x^2 = 0`

`⇔ x(4+x) = 0`

`⇔` \(\left[ \begin{array}{l}x=0\\4+x=0\end{array} \right.\) 

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

`\text{Vậy S=}` `{2; -4}`

`c)`

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

`⇔ 4-x = 2(x^2-8x+16)`

`⇔ 4-x = 2x^2-16x+32`

`⇔ 4-x-2x^2+16x-32 = 0`

`⇔ -2x^2+15x-28 = 0`

`⇔ -2x^2+7x+8x-28 = 0`

`⇔ (-2x^2+7x)+(8x-28) = 0`

`⇔ -x(2x-7)+4(2x-7) = 0`

`⇔ (4-x)(2x-7) = 0`

`⇔` \(\left[ \begin{array}{l}4-x=0\\2x-7=0\end{array} \right.\) 

`⇔` \(\left[ \begin{array}{l}x=4\\2x=7\end{array} \right.\) 

`⇔` \(\left[ \begin{array}{l}x=4\\x=\dfrac{7}{2}\end{array} \right.\) 

`\text{Vậy S=}` `{4; 7/2}`

`d)`

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

`⇔ x^3-2x^2+x-2+2x-4 = 0`

`⇔ x^3-2x^2+3x-6 = 0`

`⇔ (x^3-2x^2)+(3x-6) = 0`

`⇔ x^2(x-2)+3(x-2) = 0`

`⇔ (x^2+3)(x-2) = 0`

`\text{→Vì}` `x^2 >= 0` $\forall$ `x`

`⇒  x^2+3 >= 3` 

`\text{→Mà:}` `x^2 = -3` `\text{(loại)}`

`⇔ x-2 = 0`

`⇔ x = 2`

`\text{Vậy S=}` `{2}`