Giải phương trình `(x^2+4x-21)^2=(x+3)^4`

Đáp án:

\(\left[ \begin{array}{l}x=-15\\\left[ \begin{array}{l}x=-6\\x=1\end{array} \right. \end{array} \right.\) 

Giải thích các bước giải:

 $(x^2+4x-21)^2=(x+3)^4$

$⇔[ x^2+4x-21]^2=[(x+3)^2]^2$

$⇔$\(\left[ \begin{array}{l}x^2+4x-21=x^2+6x+9\\x^2+4x-21=-x^2-6x-9\end{array} \right.\) 

$⇔$\(\left[ \begin{array}{l}2x=-30\\2x^2+10x-12=0\end{array} \right.\) 

$⇔$\(\left[ \begin{array}{l}x=-15\\2x^2+10x-12=0\end{array} \right.\) 

$⇔$\(\left[ \begin{array}{l}x=-15\\(2x^2-2x)+(12x-12)=0\end{array} \right.\) 

$⇔$\(\left[ \begin{array}{l}x=-15\\2x.(x-1)+12.(x-1)=0\end{array} \right.\) 

$⇔$\(\left[ \begin{array}{l}x=-15\\(x-1).(2x+12)=0\end{array} \right.\) 

$⇔$\(\left[ \begin{array}{l}x=-15\\\left[ \begin{array}{l}x=-6\\x=1\end{array} \right. \end{array} \right.\)