$Bài4_{}$ $a)x^2-4=0_{}$ $⇔(x-2)(x+2)=0_{}$ $⇔x=±2_{}$ $b)_{}$ $x^2+10x+16=0_{}$ $⇔x^2+8x+2x+16=0_{}$ $⇔x.(x+8)+2(x+8)_{}$ $⇔(x+8).(x+2)=0_{}$ $⇔_{}$ \(\left[ \begin{array}{l}x+8=0\\x+2=0\end{array} \right.\)$⇔_{}$ \(\left[ \begin{array}{l}x=-8\\x=-2\end{array} \right.\) $c)_{}$ $(x^2+2x)^2-x^2-2x-2=0_{}$ $⇔x^4+4x^3+4x^2-x^2-2x-2=0_{}$ $⇔x^4+x^3+3x^3+3x^2-2x-2=0_{}$ $⇔x^3(x+1)+3x^2(x+1)-2(x=1)=0_{}$ $⇔(x+1)(x^3+3x^2-2)=0_{}$ $⇔(x+1)(x^3+x^2+2x^2-2+2x-2x)=0_{}$ $⇔(x+1)[ x^2(x+1)+2x(x+1)-2(x+1)]=0 _{}$ $⇔(x+1)(x+1)(x^2+2x-2)=0_{}$ $⇔(x+1)^2.(x^2+2x-2)=0_{}$ $⇔_{}$ \(\left[ \begin{array}{l}(x+1)^2=0\\x^2+2x-2=0\end{array} \right.\) $⇔_{}$ \(\left[ \begin{array}{l}x=-1\\x=-1±\sqrt{3}\end{array} \right.\) #Xin hay nhất ạ Reply
$Bài4_{}$
$a)x^2-4=0_{}$
$⇔(x-2)(x+2)=0_{}$
$⇔x=±2_{}$
$b)_{}$ $x^2+10x+16=0_{}$
$⇔x^2+8x+2x+16=0_{}$
$⇔x.(x+8)+2(x+8)_{}$
$⇔(x+8).(x+2)=0_{}$
$⇔_{}$ \(\left[ \begin{array}{l}x+8=0\\x+2=0\end{array} \right.\)$⇔_{}$ \(\left[ \begin{array}{l}x=-8\\x=-2\end{array} \right.\)
$c)_{}$ $(x^2+2x)^2-x^2-2x-2=0_{}$
$⇔x^4+4x^3+4x^2-x^2-2x-2=0_{}$
$⇔x^4+x^3+3x^3+3x^2-2x-2=0_{}$
$⇔x^3(x+1)+3x^2(x+1)-2(x=1)=0_{}$
$⇔(x+1)(x^3+3x^2-2)=0_{}$
$⇔(x+1)(x^3+x^2+2x^2-2+2x-2x)=0_{}$
$⇔(x+1)[ x^2(x+1)+2x(x+1)-2(x+1)]=0 _{}$
$⇔(x+1)(x+1)(x^2+2x-2)=0_{}$
$⇔(x+1)^2.(x^2+2x-2)=0_{}$
$⇔_{}$ \(\left[ \begin{array}{l}(x+1)^2=0\\x^2+2x-2=0\end{array} \right.\) $⇔_{}$ \(\left[ \begin{array}{l}x=-1\\x=-1±\sqrt{3}\end{array} \right.\)
#Xin hay nhất ạ