Question

\(x^2\)-2+x+4=0

Answer 1

\(x^2-2+x+4=-0\)

\(\Leftrightarrow x^2+x+4=0\)

\(\Rightarrow x\in\varnothing\)

Vậy phương trình vô nghiệm.

Answer 2

\(x^2-2+x+4=0\\ \left(x^2+4\right)+\left(x-2\right)=0\\ \left(x-2\right)\cdot\left(x+2\right)+\left(x-2\right)=0\\ \left(x-2\right)\left(x+2+1\right)=0\)

=> \(x-2=0\) hoặc \(x+2+1=0\)

1) \(x-2=0=>x=2\)

2) \(x+2+1=0=>x=-3\)

Vậy \(S=\left\{2;-3\right\}\)