Đáp án:Tham khảo
Giải thích các bước giải:
$ 1)x^{2}+2x-3=0$
$⇔x^{2}-x+3x-3=0$
$⇔(x^{2}-x)+(3x-3)=0$
$⇔x(x-1)+3(x-1)=0$
$⇔(x-1)(x+3)=0$
⇔\(\left[ \begin{array}{l}x-1=0\\x+3=0\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x=1\\x=-3\end{array} \right.\)
$2)x^{4}-3x^{2}+2=0$
$⇔x^{4}-x^{2}-2x^{2}+2=0$
$⇔(x^{4}-x^{2})-(2x^{2}-2)=0$
$⇔x^{2}(x^{2}-1)-2(x^{2}-1)=0$
$⇔(x²-1)(x²-2)=0$
⇔\(\left[ \begin{array}{l}x²-1=0\\x²-2=0\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x_{1}=\sqrt{1}\\x_{2}=-\sqrt{1}\\x_{1}=\sqrt{2}\\x_{2}=-\sqrt{2}\end{array} \right.\)
$3)\sqrt{x²}+3=x+1$
$⇔|x|+3=x+1$
$⇔x+3=x+1$
$⇔x-x=-3+1$
$⇔0=2$ (Vô lí)
4)
\(\left[ \begin{array}{l}x+y=2\\2x-y=1\end{array} \right.\)
⇔\(\left[ \begin{array}{l}3x=3\\y=2-x\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=1\\y=1\end{array} \right.\)
