Đáp án + Giải thích các bước giải:
`x^3=1/4x`
`<=>x^3-1/4x=0`
`<=>x(x^2-1/4)=0`
`<=>x(x-1/2)(x+1/2)=0`
`<=>`\(\left[ \begin{array}{l}x=0\\x-\dfrac{1}{2}=0\\x+\dfrac{1}{2}=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=0\\x=\dfrac{1}{2}\\x=-\dfrac{1}{2}\end{array} \right.\)
Vậy `S={0;1/2;-1/2}`
`x^3-2x^2-4x+8=0`
`<=>x^2(x-2)-4(x-2)=0`
`<=>(x-2)(x^2-4)=0`
`<=>(x-2)(x-2)(x+2)=0`
`<=>[(x-2=0),(x-2=0),(x+2=0):}`
`<=>[(x=2),(x=2),(x=-2):}<=>[(x=2),(x=-2):}`
Vậy `S={2;-2}`
`(x^2+9x)^2-36x^2=0`
`<=>(x^2+9x)^2-(6x)^2=0`
`<=>(x^2+9x-6x)(x^2+9x+6x)=0`
`<=>(x^2+3x)(x^2+15x)=0`
`<=>x(x+3)*x(x+15)=0`
`<=>x^2(x+3)(x+15)=0`
`<=>[(x^2=0),(x+3=0),(x+15=0):}`
`<=>[(x=0),(x=-3),(x=-15):}`
Vậy `S={0;-3;-15}`.
