Đáp án + Giải thích các bước giải:
`a)` `8x(x-2017)-2x+4034=0`
`<=>8x(x-2017)-2(x-2017)=0`
`<=>(x-2017)(8x-2)=0`
`<=>`\(\left[ \begin{array}{l}x-2017=0\\8x-2=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=2017\\x=\dfrac{1}{4}\end{array} \right.\)
Vậy `S={2017;1/4}`
`b)` `x/2+x^2/8=0`
`<=>(4x)/8+x^2/8=0`
`<=>(4x+x^2)/8=0`
`<=>4x+x^2=0`
`<=>x(4+x)=0`
`<=>`\(\left[ \begin{array}{l}x=0\\4+x=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=0\\x=-4\end{array} \right.\)
Vậy `S={0;-4}`
`c)` `4-x=2(x-4)^2`
`<=>-(x-4)-2(x-4)^2=0`
`<=>-(x-4)[1+2(x-4)]=0`
`<=>-(x-4)(1+2x-8)=0`
`<=>-(x-4)(2x-7)=0`
`<=>(x-4)(2x-7)=0`
`<=>`\(\left[ \begin{array}{l}x-4=0\\2x-7=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=4\\x=\dfrac{7}{2}\end{array} \right.\)
Vậy `S={4;7/2}`
`d)` `(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)(x^2+1+2)=0`
`<=>(x-2)(x^2+3)=0`
`<=>`\(\left[ \begin{array}{l}x-2=0\\x^2+3=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=0\\x=\emptyset\end{array} \right.\)
Vì `x^2>=0<=>x^2+3>=3>0`
`=>` Không tìm được giá trị của `x`
Vậy `S={2}`.
