$g$) $(x-2)^{2x+3} = (x-2)^{2x+1}$ $⇔ (x-2)^{2x+3} – (x-2)^{2x+1} = 0$ $⇔ (x-2)^{2x+1} . [(x-2)^{2} – 1] = 0$ $⇒$ \(\left[ \begin{array}{l}x=2\\x=3\\x=1\end{array} \right.\) Vậy $x$ $∈$ `{1;2;3}`. $h$) $x^2 + 2x = 0$ $⇔ x(x+2) =0$ $⇒$ \(\left[ \begin{array}{l}x=-2\\x=0\end{array} \right.\) Vậy $x$ $∈$ `{-2;0}`. $i$) $(x^4 + 0,0001)(x+1995)=0$ $⇒$ \(\left[ \begin{array}{l}x^4=-0,0001(KTM)\\x=-1995\end{array} \right.\) Vậy $x=-1995$. $k$) $(x-2) + 3x^2 – 6x = 0$ $⇔ (x-2) + 3x(x-2) = 0$ $⇔ (1+3x)(x-2) = 0$ $⇒$ \(\left[ \begin{array}{l}x=\dfrac{-1}{3}\\x=2\end{array} \right.\) Vậy $x$ $∈$ `{-1/3;2}`. Reply
$g$) $(x-2)^{2x+3} = (x-2)^{2x+1}$
$⇔ (x-2)^{2x+3} – (x-2)^{2x+1} = 0$
$⇔ (x-2)^{2x+1} . [(x-2)^{2} – 1] = 0$
$⇒$ \(\left[ \begin{array}{l}x=2\\x=3\\x=1\end{array} \right.\)
Vậy $x$ $∈$ `{1;2;3}`.
$h$) $x^2 + 2x = 0$
$⇔ x(x+2) =0$
$⇒$ \(\left[ \begin{array}{l}x=-2\\x=0\end{array} \right.\)
Vậy $x$ $∈$ `{-2;0}`.
$i$) $(x^4 + 0,0001)(x+1995)=0$
$⇒$ \(\left[ \begin{array}{l}x^4=-0,0001(KTM)\\x=-1995\end{array} \right.\)
Vậy $x=-1995$.
$k$) $(x-2) + 3x^2 – 6x = 0$
$⇔ (x-2) + 3x(x-2) = 0$
$⇔ (1+3x)(x-2) = 0$
$⇒$ \(\left[ \begin{array}{l}x=\dfrac{-1}{3}\\x=2\end{array} \right.\)
Vậy $x$ $∈$ `{-1/3;2}`.