Đáp án:
Giải thích các bước giải:
$a,2-x=2(x-2)^2$
$⇔(2-x)-2(x-2)^2=0$
$⇔(2-x)-2(2-x)^2=0$
$⇔(2-x)-2(2-x)(2-x)=0$
$⇔(2-x)-(4-2x)(2-x)=0$
$⇔(2-x)(1-4+2x)=0$
$⇔(2-x)(2x-3)=0$
⇔\(\left[ \begin{array}{l}2-x=0\\2x-3=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=2\\x=\dfrac{3}{2}\end{array} \right.\)
$b,(x-5)^4-(5-x)^2=0$
$⇔(x-5)^4-(x-5)^2=0$
$⇔(x-5)^2.[(x-5)^2-1]=0$
$⇔(x-5)^2.[(x-5-1)(x-5+1)]=0$
$⇔(x-5)^2.(x-6)(x-4)=0$
⇔\(\left[ \begin{array}{l}x-5=0\\x-6=0\\x-4=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=5\\x=6\\x=4\\\end{array} \right.\)
$c,x^2(x+1)-x(x+1)+x(x-1)=0$
$⇔(x+1)(x^2-x)+x(x-1)=0$
$⇔(x+1).x(x-1)+x(x-1)=0$
$⇔(x+1+1).x(x-1)=0$
$⇔x(x-1)(x+2)=0$
⇔\(\left[ \begin{array}{l}x=0\\x-1=0\\x+2=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=0\\x=1\\x=-2\end{array} \right.\)
