Đáp án + Giải thích các bước giải:
`a)` `4x-16x^3=0`
`<=>4x(1-4x^2)=0`
`<=>4x(1-2x)(1+2x)=0`
`<=>`\(\left[ \begin{array}{l}4x=0\\1-2x=0\\1+2x=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=0\\x=\dfrac{1}{2}\\x=-\dfrac{1}{2}\end{array} \right.\)
Vậy `S={0;1/2;-1/2}`
`b)` `x^2-8x+16=0` ( sửa đề)
`<=> (x-4)^2=0`
`<=>x-4=0`
`<=>x=4`
Vậy `S={4}`
`c)` `x^3 - 3/2x^2 + 3/4x - 1/8 = 0`
`<=> x^3-3*x^2*1/2+3*x*(1/2)^2-(1/2)^3=0`
`<=>(x-1/2)^3=0`
`<=>x-1/2=0`
`<=>x=1/2`
Vậy `S={1/2}`
`d)` `x(x-5)+2(x-5)=0<=>(x-5)(x+2)=0`
`<=>`\(\left[ \begin{array}{l}x-5=0\\x+2=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=5\\x=-2\end{array} \right.\)
Vậy `S={5;-2}`.
