`#Rùa`
Đáp án:
`x ∈{1;\sqrt{3}+1;1-\sqrt{3}}`
Giải thích các bước giải:
`x^3-3x^2=-3+1`
⇔ `x^3-3x^2=-2`
⇔ `x^3-3x^2+2=0`
⇔ `(x^3-x^2)-(2x^2-2)=0`
⇔ `x^2(x-1)-2(x^2-1)=0`
⇔ `x^2.(x-1)-2(x-1)(x+1)=0`
⇔ `(x-1)[x^2-2(x+1)=0`
⇔ `(x-1)(x^2-2x-2)=0`
⇔ \(\left[ \begin{array}{l}x-1=0\\x^2-2x-2=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=1\\(x^2-2x+1)-3=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=1\\(x-1)^2=3\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=1\\x-1=±\sqrt{3}\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=1\\x-1=\sqrt{3}\\x-1=-\sqrt{3}\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=1\\x=\sqrt{3}+1\\x=1-\sqrt{3}\end{array} \right.\)
Vậy `x ∈{1;\sqrt{3}+1;1-\sqrt{3}}`
