Đáp án + Giải thích các bước giải:
`a)` `x(x+2)-3x-6=0`
`<=>x(x+2)-3(x+2)=0`
`<=>(x+2)(x-3)=0`
`<=>`\(\left[ \begin{array}{l}x+2=0\\x-3=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=-2\\x=3\end{array} \right.\)
Vậy `S={-2;3}`
`b)` ` x^3+3x^2+3x+1-3x^2-3x=0`
`<=>(x+1)^3-3x(x+1)=0`
`<=>(x+1)[(x+1)^2-3x]=0`
`<=>(x+1)(x^2+2x+1-3x)=0`
`<=>(x+1)(x^2-x+1)=0`
`<=>x+1=0<=>x=-1`
Vì `x^2-x+1=[x^2-2*x*1/2+(1/2)^2]+3/4=(x-1/2)^2+3/4>=3/4>0`
`<=>` Không tìm được nghiệm
Vậy `S={-1}`
`c)` `8x^3-12x^2+6x-1=0`
`<=>(2x)^3-3*(2x)^2*1+3*2x*1^2-1^3=0`
`<=>(2x-1)^3=0`
`<=>2x-1=0`
`<=>x=1/2`
Vậy `S={1/2}`
`d)` `6x^4+3x^3+2x^2+x=0`
`<=>3x^3(2x+1)+x(2x+1)=0`
`<=>(2x+1)(3x^3+x)=0`
`<=>(2x+1)*x(3x^2+1)=0`
`<=>`\(\left[ \begin{array}{l}2x+1=0\\x=0\\3x^2+1=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=-\dfrac{1}{2}\\x=0\\x=\emptyset\end{array} \right.\)
Vậy `S={-1/2;0}`
`e)` `8x^3-4x^2+2x-1=0`
`<=>4x^2(2x-1)+(2x-1)=0`
`<=>(2x+1)(4x^2+1)=0`
`<=>`\(\left[ \begin{array}{l}2x+1=0\\4x^2+1=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=-\dfrac{1}{2}\\x^2=-\dfrac{1}{4}\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=-\dfrac{1}{2}\\x=\emptyset\end{array} \right.\).
Vậy `S={-1/2}`.
