Có bao nhiêu giá trị x thỏa mãn (x ² -3x +3)mũ (x ²-x-6)=1

Đáp án:

\(\left[ \begin{array}{l}x=2\\x=1\\x=-2\\x=3\end{array} \right.\) 

Giải thích các bước giải:

 $(x^2-3x+3)^{x^2-x-6}=1$

$⇒(x^2-3x+3)^{x^2+2x-3x-6}=1$

$⇔(x^2-3x+3)^{(x^2+2x)-(3x+6)}=1$

$⇔(x^2-3x+3)^{x.(x+2)-3.(x+2)}=1$

$⇔(x^2-3x+3)^{(x+2).(x-3)}=1$

$⇔$\(\left[ \begin{array}{l}x^2-3x+3=1\\(x+2).(x-3)=0\end{array} \right.\) 

$⇔$\(\left[ \begin{array}{l}x^2-2x-x+2=0\\x+2=0\\x-3=0\end{array} \right.\) 

$⇔$\(\left[ \begin{array}{l}(x-2).(x-1)=0\\x=-2\\x=3\end{array} \right.\) 

$⇔$\(\left[ \begin{array}{l}x-2=0\\x-1=0\\x=-2\\x=3\end{array} \right.\) 

$⇔$\(\left[ \begin{array}{l}x=2\\x=1\\x=-2\\x=3\end{array} \right.\) 

Đáp án:

\(\left[ \begin{array}{l}x=2\\x=-2\\x=3\\x=-1\end{array} \right.\)

Giải thích các bước giải:

`(x ² -3x +3)^(x ²-x-6)=1`

`->` \(\left[ \begin{array}{l}x^2-3x+3=1\\x^2-x-6=0\\x^2-3x+3=-1\end{array} \right.\) 

`->` \(\left[ \begin{array}{l}x^2-3x+2=0\\x^2-x-6=0\\x^2-3x+4=0\end{array} \right.\) 

`->` \(\left[ \begin{array}{l}x^2-2x-x+2=0\\x^2-3x+2x-6=0\\x^2-3x+\dfrac{9}{4}+\dfrac{7}{4}=0\end{array} \right.\) 

`->` \(\left[ \begin{array}{l}x(x-2)-(x-2)=0\\x(x-3)+2(x-3)=0\\(x-\dfrac{3}{2})^2+\dfrac{7}{4}=0(vô lý)\end{array} \right.\) 

`->` \(\left[ \begin{array}{l}(x-2)(x-1)\\(x-3)(x+2)=0\end{array} \right.\) 

`->` \(\left[ \begin{array}{l}x=2\\x=3\\x=1\\x=-2\end{array} \right.\) 

vậy \(\left[ \begin{array}{l}x=2\\x=3\\x=-2\\x=-1\end{array} \right.\) thì `(x ² -3x +3)^(x ²-x-6)=1`

$@Kate2007$