Đáp án:
`a,` `x\in{\sqrt{10}-2,-\sqrt{10}-2}`
`b,` `x\in{1,-3}`
Giải thích các bước giải:
`a,` `x^2+4x-6=0`
`⇔x^2+4x+4-10=0`
`⇔x^2+4x+4=10`
`⇔(x+2)^2=10`
⇔\(\left[ \begin{array}{l}x+2=\sqrt{10}\\x+2=-\sqrt{10}\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=\sqrt{10}-2\\x=-\sqrt{10}-2\end{array} \right.\)
Vậy `x\in{\sqrt{10}-2,-\sqrt{10}-2}`
`b,` `x^2+2x-3=0`
`⇔x^2-x+3x-3=0`
`⇔x(x-1)+3(x-1)=0`
`⇔(x-1)(x+3)=0`
⇔\(\left[ \begin{array}{l}x-1=0\\x+3=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=1\\x=-3\end{array} \right.\)
Vậy `x\in{1,-3}`
