Question

Giải phương trìnhh=)

3x(x-1)+(x+1)^2=1+3x^2

Answer 1

\(3x\left(x-1\right)+\left(x+1\right)^2=1+3x^2\)

\(\Leftrightarrow3x^2-3x+x^2+2x+1=1+3x^2\)

\(\Leftrightarrow3x^2+x^2-3x^2-3x+2x+1-1=0\)

\(\Leftrightarrow x^2-x=0\)

\(\Leftrightarrow x\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

Vậy \(S=\left\{0;1\right\}\).

Answer 2

=>3x^2-3x+x^2+2x+1=3x^2+1

=>x^2-x=0

=>x=0; x=1